$X⊗_A$ − sends finite dimensional module to a projective module $A$ a finite-dimensional $k$-algebra, $X$ projective $A ⊗_k A^{op}$-mdoule. The projectivity of $X$ as a bimodule implies that $X⊗_A$ − sends finite dimensional module to a projective module.
I am not quite familiar with bimodule projectivity. Why the tensor is projective? Thank you
 A: $M$ is projective iff $\text{Hom}(M, -)$ is exact, so we want to know why $\text{Hom}_A(X \otimes_A M, -)$ is exact. The tensor-hom adjunction gives
$$\text{Hom}_A(X \otimes_A M, -) \cong \text{Hom}_{A \otimes_k A^{op}}(X, \text{Hom}_k(M, -))$$
where, if $N$ is an $A$-module, $\text{Hom}_k(M, N)$ acquires an $A \otimes_k A^{op}$-bimodule structure by composing and precomposing by the actions of $A$ on $N$ and $M$ respectively. This means $\text{Hom}_A(X \otimes_A M, -)$ is a composition of two exact functors, namely $\text{Hom}_k(M, -)$ (I assume $k$ is a field here) and $\text{Hom}_{A \otimes_k A^{op}}(X, -)$ (by projectivity).
The following more explicit argument may also be helpful. Projective modules are retracts of free modules. If $X = \bigoplus_i A \otimes_k A^{op}$ is a free bimodule then $X \otimes_A (-) \cong \bigoplus_i A \otimes_k (-)$ is a free $A$-module (again we need to assume here that $k$ is a field). Then taking a retract of $X$ means taking a retract of this free $A$-module, which will be projective.
