Let $R$ be a finite dimensional semisimple $k$ algebra ($R$ is not necessarily commutative) and $M$ be an $R$-bimodule such that $M$ has finite dimension over $k$.
Define $M^{\ast}:=Hom_{R}(M_{R},R_{R})$, i.e $M^{\ast}$ is the dual right module of $M$. Note that $M^{\ast}$ is a k-vector space.
What is $dim_{k}M^{\ast}$? is this equal to $\dim_{k}M$ always? or more generally, what is $dim_{k}M^{\ast}$?
If $R=A\times B$ is a direct product of rings, and $M$ is an $R$-module, then there exist an $A$-module $M_1$ and a $B$-module $M_2$ such that $M\cong M_1\times M_2$ with the obvious $R$-module structure on the right and side of the isomorphism.
Moreover, it is easy to see that $\hom_R(M,N)=\hom_A(M_1,N_1)\times\hom_B(M_2,N_2)$ for all $R$-modules $M$ and $N$, with $M_1$, $M_2$, $N_1$, $N_2$ what you imagine.
It follows that the dual module $\hom_R(M,R)$ is $\hom_A(M_1,A)\times\hom_B(M_2,B)$. This reduces your question to the case where $R$ is simple artinian.
We are left with:
This question is stable under extension of the base field. If $\bar k$ is the algebraic closure of $k$, and $\bar R=R\otimes_k\bar k$ and $\bar M=M\otimes_k\bar k$, then $\bar R$ is a finite dimensional simple $\bar k$-algebra, $\bar M$ is a left $\bar R$-module, $\dim_{\bar k}\bar M=\dim_k M$, and $\dim_{\bar k}\hom_{\bar R}(\bar M,bar R)=\dim_k\hom_R(M,R)$. We can therfore suppose, without any loss, that $k$ is algebraically closed. In that case, $R=M_n(k)$ is just a matrix ring.
So we have got to
Now there is a unique simple left $R$-module $S$ up to isomorphism, so $M$ is isomorphic to a direct sum $S^r$ of some copies of $S$. Since $\hom_R(M,R)\hom_R(S^r,R)\cong\hom_R(S,R)^r$ as vector spaces, $\dim M=r\dim S$ and $\dim \hom_R(S,R)^r=r\dim\hom_R(S,R)$, w are left with
Can you answer this one?