Let $L$ be a finite field extension of $\mathbb Q_p$ and $\mathfrak o$ its ring of integers. Let $M$ be an $\mathfrak o$-module and $N$ a $\mathbb Z_p$-module. We then have the map $$ \Phi \colon \text {Hom}_{\mathbb Z_p}(M,N) \otimes_\mathbb o M \to \text{Hom}_{\mathbb Z_p}(\mathfrak o, N) \\ \,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,\,\, f \otimes m \mapsto [a \mapsto f(am)] $$ where $\text{Hom}_{\mathbb Z_p}(M,N)$ is an $\mathfrak o$-module with pointwise addition and scalar multiplication $(af)(m):=f(am)$. Same for $\text{Hom}_{\mathbb Z_p}(\mathfrak o, N)$.
Under what circumstances is the map $\Phi$ an isomorphism? Maybe when $M$ is free of finite rank over $\mathbb Z_p$?
You may also take arbitrary PIDs $A \subseteq B$ instead of $\mathbb Z_p \subseteq \mathfrak o$, if you like to think about this question in greater generality.
EDIT: A quick sanity check done in the comments (comparing ranks) shows that we need to at least assume that $M$ is free of rank one over $\mathfrak o$, in which case $\Phi$ is trivially an isomorphism. Problem solved.