Let $M\models T$ and $C\subset B\subset M$ be finite sets. Let $tp(a/d)$ divides over $C$ and $d\in acl(B)$. Can we conclude that $tp(a/B)$ divides over $C$?
Here is my attempt: Let $\phi(x,d)\in tp(a/d)$ be a formula which divides over $C$. Let $\theta(y,\bar{b})$ be a formula witnesses that $d\in acl(B)$ where $\bar{b}\in B$. Then $\phi(x,y)\wedge \theta(y,\bar{b})$ is a formula witnesses that $tp(a/B)$ divides over $C$. Is this idea true?
The answer is no: in a general theory, it is possible to have the situation that $\text{tp}(a/B)$ does not divide over $C\subseteq B$, but $\text{tp}(a/\text{acl}(B))$ divides over $C$. Even worse, it is possible that $\text{tp}(a/B)$ does not divide over $C\subseteq B$, but $\text{tp}(a/\text{acl}(C)B)$ divides over $\text{acl}(C)$.
This is a bit shocking to me - I had believed the answer was yes, and I previously wrote a detailed answer with a proof (now deleted), but the proof was incorrect. I subsequently found a counterexample, Gabe Conant provided a simplification, and I wrote it up here.
In the meantime, let me point out some positive facts:
If $\text{tp}(a/B)$ does not divide over $C\subseteq B$, then $\text{tp}(\text{acl}(Ca)/B)$ does not divide over $C$. So acl can be safely added to the left hand side in a general theory.
If $\text{tp}(a/B)$ does not fork over $C\subseteq B$, then we can add acl on both sides: $\text{tp}(\text{acl}(Ca)/\text{acl}(B))$ does not fork over $C$.
So the answer to your question is yes if you replace dividing by forking. As a consequence, the answer is yes in any theory in which forking equals dividing for complete types (e.g. any stable, simple, or o-minimal theory - or if we additionally assume that $C$ is a model, then in any $\mathrm{NTP}_2$ theory).