Let $R$ be a ring and $A, B, C$ be given $R$-modules such that we have a perfect pairing $$ \phi: A \times B \rightarrow C $$ ie $A \simeq \text{Hom}_R(B,C)$.
If $A^\prime$ is a given submodule of $A$, is there an explicit description for the corresponding $B^\prime$ such that $A^\prime \times B^\prime \rightarrow C$ is a perfect pairing?
One may as well identify $A$ with $\hom_R(B,C)$ and think of the pairing as simply function evaluation,
$*:\hom_R(B,C) \times B \to C$
where $f*b=f(b)$. Now a submodule $A'$ is just a subset of maps $f:B\to C$. To be a perfect pairing under restriction would now mean that $A'\cong \hom_R(B',C)$. Take $B'=B/\cap_{f\in A'} \ker f$. Each $f\in A'$ factors through $B'\to C$ which gives a natural map $A'\to \hom_R(B',C)$. This map is monic as $f\mapsto 0$ means $f$ is trivial on $B/\cap_{f\in A'} \ker f$ and so also on $B$; so, $f=0$. But I doubt you can force this to be surjective in all cases. Maybe there is a theorem that says otherwise, but here is an alternative reason I don't think you can.
If things are over a field $R$ (or if $A$ and $C$ are reflective $R$-modules) you can also shuffle the product to one $\circ:\hom_R(C,R)\times B\to \hom_R(A,R)$, i.e.
$(\lambda\circ b)(a) = \lambda(a*b)$
So perfect would just mean the dual $\hom_R(A,R)\cong B\otimes_R\hom_R(C,R)$. If $A'$ is any submodule of $\hom_R(A,R)$, it is hard to believe you can always choose $B'$ so that $B\otimes_R \hom_R(C,R)\cong A'$. E.g. using dimensions just choose an $A'$ of prime dimension coprime to the dimension of $C$. So perhaps the $B'$ you seek is not always going to exist.