I would guess the answer in general is hopeless, even for nice (noncommutative) algebras. For example, take $A_q := \mathbb C \langle X^{\pm 1}, Y^{\pm 1}\rangle / (XY=q^2YX)$, the quantum torus. If $q \in \mathbb C^*$ is not a root of unity, this is a simple algebra with trivial center. Let $M = P_k = \mathbb C[x^{\pm 1 }]$ as vector spaces with $1$ mapping to $m_0$ and $p_k$, respectively. Give $M$ and $P_k$ right and left $A_q$-module structures using
$f(x)m\cdot X = f(q^{-2}x) m$ and $f(x)m\cdot Y = xf(x)m$
$X\cdot f(x) p_k = xf(x)p_k$ and $Y\cdot f(x)m = q^{-k}X^k f(q^{-2})p_k$q^{-k}x^k f(q^{-2}x)p_k$. Claim: All the$P_k$are non-isomorphic, each vector space$M \otimes_{A_q} P_k$is 1-dimensional (spanned by$m\otimes p_k$), and if$q \in \mathbb C^*$is not a root of unity, then$M$and$P_k$are simple. 1 I would guess the answer in general is hopeless, even for nice (noncommutative) algebras. For example, take$A_q := \mathbb C \langle X^{\pm 1}, Y^{\pm 1}\rangle / (XY=q^2YX)$, the quantum torus. If$q \in \mathbb C^*$is not a root of unity, this is a simple algebra with trivial center. Let$M = P_k = \mathbb C[x^{\pm 1 }]$as vector spaces with$1$mapping to$m_0$and$p_k$, respectively. Give$M$and$P_k$right and left$A_q$-module structures using$f(x)m\cdot X = f(q^{-2}x) m$and$f(x)m\cdot Y = xf(x)mX\cdot f(x) p_k = xf(x)p_k$and$Y\cdot f(x)m = q^{-k}X^k f(q^{-2})p_k$. Claim: All the$P_k$are non-isomorphic, each vector space$M \otimes_{A_q} P_k$is 1-dimensional (spanned by$m\otimes p_k$), and if$q \in \mathbb C^*$is not a root of unity, then$M$and$P_k\$ are simple.