Exercise 16.1 of Jech's Set Theory says:
$B(P * \dot{Q}) = B(P) * B(\dot{Q})$.
Here, $P$ and $Q$ are (forcing) posets, and $B(P)$ and $B(Q)$ are their respective Boolean completions.
Lemma 16.3 of Jech tells us that $B(P) * B(\dot{Q})$ is a complete Boolean algebra, so it suffices to show that $P * \dot{Q}$ embeds densely into $B(P) * B(\dot{Q})$. Honestly, I'm not too sure what should the embedding be, but motivated by Lemma 16.3 I tried to define the embedding $\pi : P * \dot{Q} \to B(P) * B(\dot{Q})$ by stipulating that $\pi((p,\dot{q}))$ is the unique $\dot{c} \in B(P) * B(\dot{Q})$ such that: $$ \|\dot{c} = \dot{q}\| = p \text{ and } \|\dot{c} = -\dot{q}\| = -p $$ However, I don't think we can ensure that this $\pi$ is injective (and I am not even sure if it is a Boolean algebra homomorphism), so I'm stuck here.
Any help is appreciated.