Let $A$ and $B$ be rings and let $A\text{-mod}$ and $B\text{-mod}$ be their abelian module categories. Let $F:A\text{-mod}\to M\text{-mod}$ and $F':B\text{-mod}\to A\text{-mod}$ be functors which afford an equivalence between the two module categories (i.e. such that $F\circ F'\simeq \operatorname{id}_{A\text{-mod}}$ and $F'\circ F\simeq \operatorname{id}_{B\text{-mod}}$).
Claim Every finitely-generated $A$-module is a homomorphic image of a direct sum of copies of $P$, where $P$ is the image under $F'$ of the regular representation of $B$.
Attempt at a solution We will need to use the fact that every $A$-module is a homomorphic image of a direct sum of copies of the regular representation of $A$. I can't seem to figure out where the natural isomorphisms come into it.