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. So the embeddings 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.

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.