Ring extensions $A \subseteq B$ such that $P^{ce}=P$, for every prime ideal $P$ of $B$

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Let $A \subsetneq B$ be a ring extension, where $A$ and $B$ are commutative $k$-algebras, $k$ is a field of characteristic zero. Let $P$ be a prime ideal of $B$. Denote the contraction of $P$ to $A$ by $P^{c}=P \cap A$, and its extension to $B$ by $P^{ce}=(P^{c})^{e}=(P \cap A)^{e}=(P \cap A)B$.

I am looking for special cases in which $P^{ce}=P$.

Remarks:

(1) If $f: U \to V$ is a surjective ring homomorphism, then for every ideal of $U$ (prime or not) we have $I^{ce}=I$, see here.

(2) If $A \subset B$ is faithfully flat, then for every prime ideal $p$ of $A$, $p^{ec}=p$, see the answer to this question.

(3) Is it true that if $B$ is a finitely generated free $A$-module, then my question has a positive answer? (the condition "finitely generated" is necessary, since, for example, my question has a negative answer for $A=k$, $B=k[x,y]$).