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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.
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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 $Y\cdot f(x)m = q^{-k}X^k f(q^{-2})p_k$.

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.