Does a complete subalgebra A of a complete boolean B algebra always intersect all dense subsets of B?

I'm trying to show that if $(B, i)$ is the (BA) completion of any partial order $P$ and $A$ is a complete subalgebra of $B$, then $i^{-1}[A]$ is a complete suborder of $P$.

Pure hunch says it's true, but i'm stuck at whether a complete subalgebra $A$ of a complete boolean $B$ algebra always intersect all dense subsets of $B$.

It is not generally true that if $i:P\to B$ is the injection of the completion of $P$ and $A\subseteq B$ is a complete subalgebra of $B$ then $i^{-1}(A)$ is complete. As an example take $A=B$, which is clearly a subalgebra of $B$ but $i^{-1}(A)=P$ which clearly need not be complete.
The second assertion is also not generally true. Let $B$ be a complete boolean algebra with the property that $B$ with the top and bottom elements removed is dense. Let $A$ be the two-element set consisting of just the top and bottom elements in $B$. Clearly, $A$ is complete yet does not intersect $B-\{\top , \bot \}$, which is assumed dense.