@BugsBunny answered the original version of the question. I'll answer the new version. The algebra $B$ must be finite dimensional and semisimple under these hypothesis, and even stronger, it must be separable meaning that it remains semisimple even under base extension.

Let $B^{e}=B\otimes_k B^{op}$ be the enveloping algebra. Note that (left) $B^e$-modules equal $B$-$B$-bimodules in which the left and right actions of $k$ coincide. In particular $A$ and $B$ are $B^e$-modules.

Recall that $B$ is separable over $k$ if $B$ is a projective left $B^e$-module. This is well known to be equivalent to $B$ being finite dimensional over $k$ and for each field extension $L/K$, the algebra $L\otimes_k B$ is semisimple. All these things can be found in Pierce's book.

Now under your assumption, $A$ is a direct sum of simple $B^e$-modules that are projective. Thus $A$ is a semisimple $B^e$-module and hence the same is true for its $B^e$-submodule $B$. Moreover, if $S$ is a simple $B^e$-submodule of $B$, then it must be nontrivial under the one of the projections of $A$ onto a simple $B^e$-summand and so $S$ is isomorphic to one of the simple $B^e$-summands in $A$ by Schur's lemma. Therefore, $B$ is a direct sum of projective $B^e$-modules and hence is projective. Thus $B$ is separable over $K$ and hence finite dimensional and semisimple (even after base change).

So your desired situation cannot occur if $B$ is infinite dimensional over $k$.

If you drop the projective hypothesis you could take $B$ a finite direct product of simple $k$-algebras at least one of which is infinite dimensional and take $A=B$ and $A$ will be a finite direct sum of simple $B$-bimodules. You can even make $B$ finitely presented as a $k$-algebra.

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