Suppose $\mathbb{P}$ is a regular suborder of the separative partial order $\mathbb{Q}$ (see below for definitions). Must there always exist some complete boolean algebra $\mathbb{B}$ such that:


1. $\mathbb{P}$ is a dense suborder of $\mathbb{B}$ (so $\mathbb{B}$ is the boolean completion of $\mathbb{P}$)
2. $\mathbb{B}$ is a subalgebra of the boolean completion $\text{ro}(\mathbb{Q})$ of $\mathbb{Q}$ (here we are viewing $\mathbb{Q}$ as a dense suborder of its boolean completion $\text{ro}(\mathbb{Q})$)
3. $\mathbb{B}$ is also regular in $\text{ro}(\mathbb{Q})$


It is a standard fact that one can lift any regular embedding of separative posets to a regular embedding of the boolean completions. The question is basically whether this lifting can also be viewed as an identity map in a natural way, if the original map was the identity. I suspect the answer is no, and must be known, but do not know of a counterexample. 


Definition: Let $\mathbb{P}$ and $\mathbb{Q}$ be separative partial orders. A map $e: \mathbb{P} \to \mathbb{Q}$ is called a regular embedding (or complete embedding) iff it is order and incompatibility preserving, and whenever $A$ is a maximal antichain in $\mathbb{P}$ then $e[A]$ is maximal in $\mathbb{Q}$. The latter is equivalent to saying that every $q \in \mathbb{Q}$ has an $e$-reduction in $\mathbb{P}$; i.e. there is some $p \in \mathbb{P}$ such that whenever $p' \le p$ then $e(p') \parallel q$.



completions of regular suborders
$\newcommand\P{\mathbb{P}}\newcommand\Q{\mathbb{Q}}
\newcommand\Q{\mathbb{Q}}\newcommand\B{\mathbb{B}}\newcommand\Z{\mathbb{Z}}
\newcommand\RO{\text{RO}}$Unless I am mistaken (and please correct me if I am, since these issues are sometimes confusing), I believe the answer is yes. Let $\B$
consist of the elements of $\RO(\Q)$ that are the join of a subset
of $\P$. This is the same as the elements in $\RO(\Q)$ that are
the join of an antichain in $\P$, since if $a=\bigvee X$ with
$X\subset\P$, then let $A\subset\P$ be an antichain in $\P$ that
is maximal consisting of elements pointwise below elements of $X$, and note that $a=\bigvee A$.
Clearly, $\B$ is closed under arbitrary joins. Also, $\B$ is
closed under complements, since if $a=\bigvee A$ for an antichain
$A\subset\P$, then we may extend $A$ to a
maximal antichain $A\sqcup B$ in $\P$, which is also maximal in
$\Q$, and so $\neg a=\bigvee B$. Thus, $\B$ is the complete
subalgebra of $\RO(\Q)$ generated by $\P$.


Clearly $\P$ is dense in $\B$ and $\B$ is a complete subalgebra of
$\RO(\Q)$, and so $\B$ is as desired.


Lastly, although in your question you had assumed only that $\Q$
is separative and not that $\P$ is separative, let me remark that
actually it follows from the other assumptions that $\P$ must be
separative. To see this, suppose $x\not\leq y$ in $\P$, but there
is no $z\in\P$ with $z\leq x$ and $z\perp y$. So we may find a
maximal antichain $A\cup\{y\}$ in $\P$ where every element of $A$
is incompatible with $x$. This antichain cannot be maximal in
$\Q$, however, since being separative, $\Q$ has an element $z\leq
x$ with $z\perp y$, and this $z$ is therefore incompatible with
every element of $A\cup\{y\}$, violating that $\P$ is regular in
$\Q$.


