Let $(A,H,D)$ be a spectral triple and let $B$ be an algebra which is Morita equivalent to $A$. Then there exists a finitely generated, projective $A$ module $E$ such that $B=End\_A(E)$. Endow $E$ with $A$-valued inner product $(\cdot,\cdot)$ and form a Hilbert space $H':=E \otimes\_A H$ with the inner product given by 
$$\langle x \otimes \xi, y \otimes \eta \rangle\_{H'}:=\langle (x,y)\xi,\eta \rangle\_H$$
One would like impose the structure of a spectral triple on $B$: one naive choice is to put $D'(x \otimes \xi)=x \otimes D\xi$ however this map is not well defined since $H'$ is a tensor product *over $A$* 
In order to fix this one uses a connection $\nabla$ i.e. a map $\nabla:E \to E \otimes\_A \Omega\_D^1$ where $\Omega^1\_d=\{\sum\_{j=1}^N a\_jdb\_j: a\_j,b\_j \in A, N \in \mathbb{N}\}$ and $db:=[D,b]$. This map has to satisfy $\nabla(xa)=\nabla(x)a+x\otimes da$ (note that due to Jacobi idenity for commutators the space $\Omega\_D^1$ is in fact $A-A$-bimodule thus $E \otimes\_A \Omega^1\_D$ is a right $A$-module). Having fixed a connection one defines 
$$D'(x \otimes \xi)=x\otimes D\xi + \nabla(x)\xi$$ 
(where $\nabla(x)\xi$ is understood as follows: if $\nabla(x)=\sum\_j x\_j \otimes \omega\_j$ then $\nabla(x)\xi=\sum\_j x\_j \otimes \omega\_j(\xi)$ which is fine since each $\omega\_j \in \Omega^1\_D$ still acts on $H$).
The natural action of $B$ on $H'$ is as follows $b(x \otimes \xi)bx \otimes \xi$ which is fine since $b \in B=End\_A(E)$. 



> 
> How to show that for each $b \in B$ the commutator $[D',b]$ (extends to a/) is a bounded operator on $H'$?
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Simple computation shows that this commutator evaluated on $x \otimes \xi$ is equal to $\nabla(bx)\xi-b(\nabla x(\xi))$ and I don't see how to obtain that this operator is bounded-do we have to assume something on $\nabla$ (some sort of compatibility with the $B$-action)? 



Dirac operator on a Morita equivalent algebra
Your question is entirely covered by Section 2 of [Brain–Mesland–Van Suijlekom](https://arxiv.org/abs/1306.1951), but the fgp case is simple enough to ultimately boil down to folklore proved by [Chakraborty–Mathai](https://arxiv.org/abs/0804.3232). Let me summarise what happens, while incorporating some technical simplifications from [Blecher–Kaad–Mesland](https://arxiv.org/abs/1703.10063). For convenience, let me take $A$ and $B$ to be unital $C^\ast$-algebras and let $\Omega\_D$ be the unital $C^\ast$-subalgebra of $B(H)$ generated by $A$ and $[D,\mathcal{A}]$.


1. Let $\mathcal{A} := \{a \in A \mid a \operatorname{Dom}(D) \subset \operatorname{Dom}(D),\, [D,a] \in B(H)\}$ be given the Lipschitz norm $\|a\| := \|a\| + \|[D,a]\|$. Then $\mathcal{A}$ defines an involutive operator algebra and the inclusion $\mathcal{A} \hookrightarrow A$ is completely bounded with dense range that is closed under the holomorphic functional calculus; in particular, it follows that $M\_N(\mathcal{A})$ is dense in $M\_N(A)$ and closed under the holomorphic functional calculus for any $N \in \mathbb{N}$. You should think of $\mathcal{A}$ as defining a Lipschitz structure on the NC space $A$.
2. Let $\mathcal{E}$ be a dense subspace of $E$ satisfying $\mathcal{E} \cdot \mathcal{A} \subset \mathcal{E}$ and $(\mathcal{E},\mathcal{E})\_A \subset \mathcal{A}$. By part 1 and the properties of $\mathcal{E}$, we can find $A$-module generators $\{\xi\_1,\dotsc,\xi\_N\} \subset \mathcal{E}$ for $E$, such that 
$$
 \forall e \in E, \quad e = \sum\_{i=1}^N \xi\_i \cdot (\xi\_i,e)\_A.
$$
Thus, if $p := \left((\xi\_i,\xi\_j)\_A\right)\_{i,j=1}^N \in M\_N(\mathcal{A})$, then $e \mapsto \left((\xi\_i,e)\_A\right)\_{i=1}^N$ defines an isomorphism $E \cong pA^N$ of Hilbert $A$-modules that restricts to an isomorphism $\mathcal{E} \cong p\mathcal{A}^N$ of pre-Hilbert $\mathcal{A}$-modules; this now makes $\mathcal{E}$ into a (finitely generated) projective operator module over $\mathcal{A}$ in a manner that depends on the choice of $\{\xi\_1,\dotsc,\xi\_N\}$ only up to completely bounded isomorphism. You should think of $\mathcal{E}$ as defining a Lipschitz structure on the NC vector bundle $E$; since $E$ is fgp, this Lipschitz structure is canonically induced by the choice of Lipschitz structure on $A$. Note that such $\mathcal{E}$ exists: given an isomorphism $E \cong p\_0 A^N$ of Hilbert $A$-modules for $p\_0 \in M\_N(A)$ an orthogonal projection, you can use the properties of $\mathcal{A}$ to find an orthogonal projection $p \in M\_N(\mathcal{A})$, such that $E \cong p A^N$, in which case, you can take $\mathcal{E}$ to be the pre-image of $p\mathcal{A}^N$ and $\{\xi\_1,\dotsc,\xi\_N\}$ to be given by the pre-images in $E$ of the columns of $p$.
3. Let $\mathcal{B} := \{b \in B \mid b \cdot \mathcal{E} \subset \mathcal{E}\}.$ Since
$$ 
 \mathcal{B} = \left\{b \in B \mid \left((\xi\_i,b\xi\_j)\_A\right)\_{i,j=1}^N \in M\_N(\mathcal{A})\right\},
$$
it follows that $\mathcal{B}$ is $\ast$-closed and that the $\ast$-isomorphism $B \cong p M\_N(A) p$ induced by the Hilbert $A$-module isomorphism $E \cong p A^N$ restricts to a $\ast$-isomorphism $\mathcal{B} \cong p M\_N(\mathcal{A}) p$. Thus, $\mathcal{B}$ can be topologised as a closed $\ast$-subalgebra of the involutive operator algebra $M\_N(\mathcal{A})$, thereby making $\mathcal{E}$ into a Lipschitz $(\mathcal{B},\mathcal{A})$-correspondence. In particular, you can view $\mathcal{B}$ as defining a Lipschitz structure on the NC space $B$ compatible with the Lipschitz structure $\mathcal{A}$ on the NC space $A$.
4. Let $\nabla : \mathcal{E} \to E \hat\otimes\_A \Omega\_D$ be a *Hermitian connection*, i.e., a $\mathbb{C}$-linear map satisfying 
$$
 \forall e \in \mathcal{E},\, \forall a \in \mathcal{A}, \quad \nabla(ea) = \nabla(e)a + e \hat\otimes [D,a],\\
 \forall e\_1,e\_2 \in \mathcal{E}, \quad (\nabla(e\_1),e\_2 \hat\otimes 1)\_{\Omega\_D} +(e\_1 \hat\otimes 1,\nabla(e\_2))\_{\Omega\_D} = [D,(e\_1,e\_2)\_A];
$$
for example, the *Graßmann connection* $\nabla\_0$ induced by the frame $\{\xi\_1,\dotsc,\xi\_N\}$ is defined by
$$
 \forall e \in \mathcal{E}, \quad \nabla\_0(e) := \sum\_{i=1}^N \xi\_i \hat\otimes [D,(\xi\_i,e)\_A],
$$
and indeed, if $\nabla$ is any other Hermitian connection, then $\nabla = \nabla\_0 + \omega$ for $\omega \in B(E,E \hat\otimes\_A \Omega\_D)$ defined by
$$
 \forall e \in \mathcal{E}, \quad \omega(e) := \sum\_{i=1}^N \nabla(\xi\_i)(\xi\_i,e)\_A.
$$
Given a Hermitian connection $\nabla$, define $1 \hat\otimes\_\nabla D : \mathcal{E} \otimes^{\mathrm{alg}}\_{\mathcal{A}} \operatorname{Dom}(D) \to E \hat\otimes\_A H$ by
$$
 \forall e \in \mathcal{E}, \, \forall h \in H, \quad 1 \hat\otimes\_\nabla D(e \hat\otimes h) := \nabla(e)h + e \hat\otimes Dh = \omega(e)h + 1 \hat\otimes\_{\nabla\_0}D(e \hat\otimes h).
$$
Then, by Section 2 of Chakraborty–Mathai, the operator $1 \hat\otimes\_\nabla D$ is essentially self-adjoint and defines a spectral triple $(B,E \hat\otimes\_A H,1 \hat\otimes\_\nabla D)$ with Lipschitz algebra $\mathcal{B}$. In particular, by a direct computation,
$$
 \forall b \in \mathcal{B}, \, \forall e \in E, \, \forall h \in H, \quad [1 \hat\otimes\_{\nabla\_0} D,b](e \hat\otimes h) = \sum\_{i,j=1}^N \xi\_i \hat\otimes [D,(\xi\_i,b\xi\_j)\_A](\xi\_j,e)\_A h;
$$
so that $1 \hat\otimes\_\nabla D$, which is a bounded perturbation of $1 \hat\otimes\_{\nabla\_0} D$, also has bounded commutators with $\mathcal{B}$.


To conclude, the missing ingredient was a Lipschitz structure on the fgp $A$-module $E$, i.e., a dense subspace $\mathcal{E}$ of $E$, such that $\mathcal{E} \cdot \mathcal{A} \subset \mathcal{E}$ and $(\mathcal{E},\mathcal{E})\_A \subset \mathcal{A}$. Because $E$ is fgp, this suffices to pick out a dense $\ast$-subalgebra $\mathcal{B}$ of Lipschitz elements of $B$ and to pin down any Hermitian connection $\nabla$ on $E$ with domain $\mathcal{E}$ up to bounded perturbation, so that $\mathcal{B}$ will necessarily have bounded commutators with $1 \hat\otimes\_\nabla D$; indeed, by using a normalised tight frame for $E$ consisting of vectors in $\mathcal{E}$, you can boil everything down to the compatibility of $\mathcal{A}$ with $D$.


