Let $A$ and $B$ be two C$^{\*}$-algebras and suppose we have a non-zero projection $p\in A\otimes B$. (We can assume $A$ is nuclear, so that there is only one possible tensor product.) 


Does there exist a choice of elements $a\_{1},\ldots,a\_{n}\in A$ and $b\_{1},\ldots,b\_{n}\in B$ such that:


1. $\left\|\left(\sum\_{i=1}^{n}a\_{i}\otimes
 b\_{i}\right)-p\right\|<\frac{1}{2}$;
2. If $\pi$ is a non-zero irreducible representation of $A$ and $(\pi\otimes\operatorname{id})(p)=0$, then $\pi\left(\sum\_{i=1}^{n}|a\_{i}|\right)=0$?


All I can deduce is that for a given choice of $a\_{i}$'s and $b\_{i}$'s, and and irreducible representation $\pi$ of $A$, we have
$$
\left\|\sum\_{i=1}^{n}\pi(a\_{i})\otimes b\_{i}\right\|=\left\|(\pi\otimes \operatorname{id})\left(\left(\sum\_{i=1}^{n}a\_{i}\otimes b\_{i}\right)-p\right)\right\|\leq \left\|\left(\sum\_{i=1}^{n}a\_{i}\otimes b\_{i}\right)-p\right\|<\frac{1}{2},
$$
provided that $(\pi\otimes\operatorname{id})(p)=0$.



Approximating a projection by a sum of elementary tensors with a certain property
Yes, this is possible, assuming that $A$ (or $B$) is nuclear. The same argument below (using that exact $C^\ast$-algebras are locally reflexive) also works if $A$ or $B$ is exact and the tensor product is the spatial (aka minimal) tensor product.


The role of nuclearity is that any two-sided closed ideal $J \subseteq A\otimes B$ is the closed linear span of all tensor products $I\_A \otimes I\_B$ of two-sided closed contained in $J$, see Corollary 9.4.6 in Brown and Ozawa's book. Now, let $J\_A \subseteq A$ be the intersection of all two-sided, closed ideals $J \subseteq A$ such that $p \in J \otimes B$ and let $I:= \bigcap J \otimes B$ where the intersection is indexed by such $J$. Clearly $J\_A \otimes B \subseteq I$. For the converse (which is not as trivial as it a priori looks), take $I\_A \otimes I\_B \subseteq I$ where $I\_A$ and $I\_B$ are two-sided closed ideals in $A$ and $B$ respectively. Then $I\_A \subseteq J\_A$ and $I\_B \subseteq B$, so $I\_A \otimes I\_B \subseteq J\_A \otimes B$. By nuclearity of $A$, $I$ is the closed linear span of all such $I\_A\otimes I\_B$, so $I = J\_A \otimes B$. In particular, $p\in J\_A \otimes B$.


Now, let $a\_1,\dots, a\_n \in J\_A$ and $b\_1,\dots, b\_n \in B$ such that
\begin{equation}
\| \sum\_{i=1}^n a\_i \otimes b\_i - p \| < 1/2.
\end{equation}
Let $\pi \colon A \to \mathcal B(\mathcal H)$ be an irreducible representation such that $(\pi \otimes \mathrm{id}\_B)(p) = 0$. The image of $(\pi \otimes \mathrm{id}\_B)$ is canonically $\pi(A) \otimes B$, so by nuclearity of $A$ (or $B$) and exactness of maximal tensor products, it follows that 
\begin{equation}
0 \to (\ker \pi)\otimes B \to A \otimes B \to \pi(A) \otimes B \to 0
\end{equation}
is exact, so the kernel of $(\pi \otimes \mathrm{id}\_B)$ is $(\ker \pi )\otimes B$. Hence $p \in (\ker \pi)\otimes B$ so $a\_1,\dots, a\_n \in J\_A \subseteq \ker \pi$ by construction of $J\_A$.


If $A$ or $B$ is exact, we get $\ker (\pi \otimes \mathrm{id}\_B)$ either by using that $A$ is locally reflexive or $B$ is exact to get that
\begin{equation}
0 \to (\ker \pi) \otimes\_{\min{}} B \to A \otimes\_{\min{}} B \to \pi(A) \otimes\_{\min{}} B \to 0
\end{equation}
is exact.


