Fix a finitely generated left $R$-module $_RP$ that is in $\mathcal{C}$ and projective as an object of $\mathcal{C}$, and a finite dimensional left $S$-module $_SC$.

As noted in the question, for any $R$-$S$-bimodule $_RM_S$ there is an obvious map $$\Phi_M:\operatorname{Hom}_R(P,M)\otimes_SC\to\operatorname{Hom}_R(P,M\otimes_SC),$$ and it is natural in $M$.

If $M$ is of the form $X\otimes_\mathbb{C}S$ for $_RX$ a left $R$-module, then $\Phi_M$ is an isomorphism, using the fact that $_RP$ is finitely generated as a left $R$-module, so that the natural map $$\operatorname{Hom}_R(P,X)\otimes_\mathbb{C}V\to \operatorname{Hom}_R(P,X\otimes_\mathbb{C}V)$$ is an isomorphism for any vector space $V$.

The standard projective bimodule presentation $$S\otimes_\mathbb{C}S\otimes_\mathbb{C}S\to S\otimes_\mathbb{C}S\to S\to0$$ of $S$ induces an exact sequence $$\tag{*} B\otimes_\mathbb{C}S\otimes_\mathbb{C}S\to B\otimes_\mathbb{C}S\to B\to0,$$ which is split as an exact sequence of left $R$-modules, so that applying the functor $M\mapsto\operatorname{Hom}_R(P,M)\otimes_S C$ to it preserves exactness.

Also, applying $-\otimes_SC$ to the exact sequence $(*)$ gives an exact sequence $$\tag{**} B\otimes_\mathbb{C}S\otimes_\mathbb{C}C\to B\otimes_\mathbb{C}C\to B\otimes_SC\to0. $$

As a sequence of left $R$-modules, the first term is a (possibly infinite) direct sum $\bigoplus_{i\in I}B$ of copies of $B$ and the second is (since $C$ is finite dimensional) a finite direct sum of copies of $B$.

This is the filtered coproduct, over finite subsets $J\subseteq I$, of exact sequences of the form $$\tag{***} \bigoplus_{j\in J}B\to B\otimes_\mathbb{C}C\to U_J\to 0,$$ where the first two terms are in $\mathbb{C}$ since $_RB$ is in $\mathcal{C}$, and since $\mathcal{C}$ is an abelian subcategory, the third term is also in $\mathcal{C}$. Therefore $\operatorname{Hom}_R(P,-)$ preserves exactness of the sequences $(***)$ since $P$ is projective in $\mathcal{C}$. Since $P$ is finitely generated as an $R$-module, $\operatorname{Hom}_R(P,-)$ also preserves filtered colimits, and so applying $\operatorname{Hom}_R(P,-)$ to $(**)$ also gives an exact sequence.

So applying $$\Phi:\operatorname{Hom}_R(P,-)\otimes_SC\to\operatorname{Hom}_R(P,-\otimes_SC)$$ to the sequence $(*)$ gives a map of exact sequences which is an isomorphism on the first two terms and therefore also on the third term.

So $\Phi_B$ is an isomorphism.

This doesn't seem to require $B\otimes_SC$ to be finite dimensional.

Fix a finitely generated left $R$-module $_RP$ that is in $\mathcal{C}$ and projective as an object of $\mathcal{C}$, and a finite dimensional left $S$-module $_SC$.

As noted in the question, for any $R$-$S$-bimodule $_RM_S$ there is an obvious map $$\Phi_M:\operatorname{Hom}_R(P,M)\otimes_SC\to\operatorname{Hom}_R(P,M\otimes_SC),$$ and it is natural in $M$.

If $M$ is of the form $X\otimes_\mathbb{C}S$ for $_RX$ a left $R$-module, then $\Phi_M$ is an isomorphism, using the fact that $_RP$ is finitely generated as a left $R$-module, so that the natural map $$\operatorname{Hom}_R(P,X)\otimes_\mathbb{C}V\to \operatorname{Hom}_R(P,X\otimes_\mathbb{C}V)$$ is an isomorphism for any vector space $V$.

The standard projective bimodule presentation $$S\otimes_\mathbb{C}S\otimes_\mathbb{C}S\to S\otimes_\mathbb{C}S\to S\to0$$ of $S$ induces an exact sequence $$\tag{*} B\otimes_\mathbb{C}S\otimes_\mathbb{C}S\to B\otimes_\mathbb{C}S\to B\to0,$$ which is split as an exact sequence of left $R$-modules, so that applying the functor $M\mapsto\operatorname{Hom}_R(P,M)\otimes_S C$ to it preserves exactness.

Also, applying $-\otimes_SC$ to the exact sequence $(*)$ gives an exact sequence $$\tag{**} B\otimes_\mathbb{C}S\otimes_\mathbb{C}C\to B\otimes_\mathbb{C}C\to B\otimes_SC\to0. $$

As a sequence of left $R$-modules, the first term is a (possibly infinite) direct sum $\bigoplus_{i\in I}B$ of copies of $B$ and the second is (since $C$ is finite dimensional) a finite direct sum of copies of $B$.

This is the filtered colimit, over finite subsets $J\subseteq I$, of exact sequences of the form $$\tag{***} \bigoplus_{j\in J}B\to B\otimes_\mathbb{C}C\to U_J\to 0,$$ where the first two terms are in $\mathbb{C}$ since $_RB$ is in $\mathcal{C}$, and since $\mathcal{C}$ is an abelian subcategory, the third term is also in $\mathcal{C}$. Therefore $\operatorname{Hom}_R(P,-)$ preserves exactness of the sequences $(***)$ since $P$ is projective in $\mathcal{C}$. Since $P$ is finitely generated as an $R$-module, $\operatorname{Hom}_R(P,-)$ also preserves filtered colimits, and so applying $\operatorname{Hom}_R(P,-)$ to $(**)$ also gives an exact sequence.

So applying $$\Phi:\operatorname{Hom}_R(P,-)\otimes_SC\to\operatorname{Hom}_R(P,-\otimes_SC)$$ to the sequence $(*)$ gives a map of exact sequences which is an isomorphism on the first two terms and therefore also on the third term.

So $\Phi_B$ is an isomorphism.

This doesn't seem to require $B\otimes_SC$ to be finite dimensional.

Fix a finitely generated left $R$-module $_RP$ that is in $\mathcal{C}$ and projective as an object of $\mathcal{C}$, and a finite dimensional left $S$-module $_SC$.

As noted in the question, for an $R$-$S$-bimodule $_RB_S$ there is an obvious map $$\Phi_B:\operatorname{Hom}_R(P,B)\otimes_SC\to\operatorname{Hom}_R(P,B\otimes_SC),$$ and it is natural in $B$.

Let $\mathcal{B}$ be the category of $R$-$S$-bimodules $_RB_S$ such that $_RB$ is in $\mathcal{C}$. Since $\mathcal{C}$ is an abelian subcategory of the category of left $R$-modules, $\mathcal{B}$ is an abelian subcategory of the category of $R$-$S$-bimodules.

Every object $B$ of $\mathcal{B}$ is a quotient of an object of the form $X\otimes_\mathbb{C}S$, where $_RX$ is an object of $\mathcal{C}$, since there is an epimorphism $b\otimes s\mapsto bs$ $$_RB\otimes_\mathbb{C}S_S\to B$$ of $R$-$S$-bimodules.

So for any $B$, there is an exact sequence $$X'\otimes_\mathbb{C}S\to X\otimes_\mathbb{C}S\to B\to0,$$ where $X$ and $X'$ are in $\mathcal{C}$, inducing an exact sequence $$X'\otimes_\mathbb{C}S\otimes_SC\to X\otimes_\mathbb{C}S\otimes_SC\to B\otimes_SC\to0.$$ Since the first two terms are in $\mathcal{C}$, which is an abelian subcategory of the category of left $R$-modules, so is $B\otimes_SC$.

Let $\mathcal{B}'$ be the full subcategory of $\mathcal{B}$ consisting of the objects $B$ for which $\Phi_B$ is an isomorphism.

Since the functors $B\mapsto\operatorname{Hom}_R(P,B)\otimes_SC$ and $B\mapsto\operatorname{Hom}_R(P,B\otimes_SC)$ are both right exact functors from $\mathcal{B}$ to vector spaces (since $P$ is projective in $\mathcal{C}$), the category $\mathcal{B}'$ is closed under cokernels.

Also, $\mathcal{B}'$ contains all objects of the form $X\otimes_\mathbb{C}S$ for $X$ in $\mathcal{C}$, using the fact that $_RP$ is finitely generated as a left $R$-module, so that the natural map $$\operatorname{Hom}_R(P,X)\otimes_\mathbb{C}S\to\operatorname{Hom}_R(P,X\otimes_\mathbb{C}S)$$ is an isomorphism.

So, since we know that every object of $\mathcal{B}$ is the cokernel of a map between such objects, $\mathcal{B}'=\mathcal{B}$.

Fix a finitely generated left $R$-module $_RP$ that is in $\mathcal{C}$ and projective as an object of $\mathcal{C}$, and a finite dimensional left $S$-module $_SC$.

As noted in the question, for any $R$-$S$-bimodule $_RM_S$ there is an obvious map $$\Phi_M:\operatorname{Hom}_R(P,M)\otimes_SC\to\operatorname{Hom}_R(P,M\otimes_SC),$$ and it is natural in $M$.

If $M$ is of the form $X\otimes_\mathbb{C}S$ for $_RX$ a left $R$-module, then $\Phi_M$ is an isomorphism, using the fact that $_RP$ is finitely generated as a left $R$-module, so that the natural map $$\operatorname{Hom}_R(P,X)\otimes_\mathbb{C}V\to \operatorname{Hom}_R(P,X\otimes_\mathbb{C}V)$$ is an isomorphism for any vector space $V$.

The standard projective bimodule presentation $$S\otimes_\mathbb{C}S\otimes_\mathbb{C}S\to S\otimes_\mathbb{C}S\to S\to0$$ of $S$ induces an exact sequence $$\tag{*} B\otimes_\mathbb{C}S\otimes_\mathbb{C}S\to B\otimes_\mathbb{C}S\to B\to0,$$ which is split as an exact sequence of left $R$-modules, so that applying the functor $M\mapsto\operatorname{Hom}_R(P,M)\otimes_S C$ to it preserves exactness.

Also, applying $-\otimes_SC$ to the exact sequence $(*)$ gives an exact sequence $$\tag{**} B\otimes_\mathbb{C}S\otimes_\mathbb{C}C\to B\otimes_\mathbb{C}C\to B\otimes_SC\to0. $$

As a sequence of left $R$-modules, the first term is a (possibly infinite) direct sum $\bigoplus_{i\in I}B$ of copies of $B$ and the second is (since $C$ is finite dimensional) a finite direct sum of copies of $B$.

This is the filtered coproduct, over finite subsets $J\subseteq I$, of exact sequences of the form $$\tag{***} \bigoplus_{j\in J}B\to B\otimes_\mathbb{C}C\to U_J\to 0,$$ where the first two terms are in $\mathbb{C}$ since $_RB$ is in $\mathcal{C}$, and since $\mathcal{C}$ is an abelian subcategory, the third term is also in $\mathcal{C}$. Therefore $\operatorname{Hom}_R(P,-)$ preserves exactness of the sequences $(***)$ since $P$ is projective in $\mathcal{C}$. Since $P$ is finitely generated as an $R$-module, $\operatorname{Hom}_R(P,-)$ also preserves filtered colimits, and so applying $\operatorname{Hom}_R(P,-)$ to $(**)$ also gives an exact sequence.

So applying $$\Phi:\operatorname{Hom}_R(P,-)\otimes_SC\to\operatorname{Hom}_R(P,-\otimes_SC)$$ to the sequence $(*)$ gives a map of exact sequences which is an isomorphism on the first two terms and therefore also on the third term.

So $\Phi_B$ is an isomorphism.

This doesn't seem to require $B\otimes_SC$ to be finite dimensional.