Suppose $\mathscr{A}$ is a complete Boolean algebra (assume it is c.c.c. if you wish). Is there any, say, *canonical* embedding $\mathscr{A}\subseteq \mathscr{B}$ into a complete Boolean algebra which is weakly countably distributive ($(\omega, \omega)$-distributive)? Note that I do not put any extra assumptions on $\mathscr{B}$ but I would like to have it as small as possible, whence $\mathscr{P}(\kappa)$ is not a good choice in general.


One way to produce such an algebra is taking 


$$\mathscr{B}=\mbox{Clop}(\mbox{Gelfand\_spectrum}(C(\mbox{Stone}(\mathscr{A}))^{\*\*})$$


provided the Gelfand spectrum of $C(\mbox{Stone}(\mathscr{A}))^{\*\*}$ is c.c.c. This is however, in a certain vague sense, much larger that $\mathscr{A}$, since it is not $\mathscr{A}$ in case $\mathscr{A}$ is already c.c.c. complete and weakly countably distributive.



Extending BAs to weakly countably distributive algebras.
Let me make a few observations.


First, although you have insisted that the Boolean algebras be complete, there can be no *complete* embedding of $\mathscr{A}$ into a distributive $\mathscr{B}$, if $\mathscr{A}$ did not already exhibit that degree of distributivity, since a complete embedding would preserve any counterexample to distributivity. Or, from the forcing perspective, if $\mathscr{B}$ doesn't add reals and $\mathscr{A}$ is a complete subalgebra, then $\mathscr{A}$ can't add reals either. So the embeddings here should be merely injective Boolean algebra homomorphisms, rather than complete Boolean algebra homomorphisms. 


Second, there can be no *dense* embedding, that is, an embedding whose image is dense in $\mathscr{B}$, if $\mathscr{A}$ was not already distributive, since a dense embedding would make the two Boolean algebras equivalent as forcing notions, and $(\omega,\omega)$-distributivity is preserved under forcing equivalence, since it is equivalent to the forcing extension not adding reals. 


Third, as I mentioned in the comments, every Boolean algebra $\mathscr{A}$ maps canonically into a fully distributive complete Boolean algebra. This is just because every Boolean algebra $\mathscr{A}$ can be realized as a field of sets, via the Stone space, by mapping every $a\in \mathscr{A}$ to the set of ultrafilters containing $a$. Thus, we canonically map $\mathscr{A}$ into the power set algebra $P(S)$, where $S$ is the set of ultrafilters on $\mathscr{A}$. Since the power set algebra is atomic, it is $(\kappa,\lambda)$-distributive for every $\kappa,\lambda$, and so we've got the desired embedding. 


But, as you note, $P(S)$ is generally much larger than $\mathscr{A}$. For example, there are generally $2^{2^{\kappa}}$ many ultrafilters on an infinite cardinal $\kappa$. 


If one gives up on the completeness of the target algebra (and relaxes the canonicity), then one can achieve a target the same size as $\mathscr{A}$. That is, every Boolean algebra $\mathscr{A}$ embeds into a fully distributive Boolean algebra $\mathscr{B}$---that is, $(\kappa,\lambda)$-distributive for every $\kappa,\lambda$---of the same cardinality as $\mathscr{A}$. To see this, simply let $\mathscr{B}$ be the elementary substructure of $P(S)$ generated by the image of $\mathscr{A}$ under the canonical map mentioned above. Thus, we map $\mathscr{A}$ into $\mathscr{B}$, which is an atomic Boolean algebra, since this is expressible in the theory of the Boolean algebra, and every atomic Boolean algebra is fully distributive. But generally $\mathscr{B}$ will have $|\mathscr{A}|$ many atoms, and so its completion will be a power set algebra of size $2^{|\mathscr{A}|}$. The map in this case is not necessarily canonical, since we used a Skolem function to find the elementary substructure.


