Ring extensions $A \subseteq B$ such that $P^{ce}=P$, for every prime ideal $P$ of $B$ 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 $A=k$, $B=k[x,y]$).
I know it has been a while since the question was posted, anyway (if I'm right) the following should be an example of $\mathfrak p^{ce}=\mathfrak p.$

Let $f\colon A\longrightarrow B$ be an injective, flat ring homomorphism. Let $\mathfrak p\in\mathsf{Spec}(B)$ and $\mathfrak q:=f^{-1}(\mathfrak p)$. I claim that $$\mathfrak qB=\mathfrak p.$$

The ring map $B/\mathfrak qB\longrightarrow B/\mathfrak p$ is onto. We show that it is injective too. As $f$ is flat and $A/\mathfrak q\longrightarrow B/\mathfrak p$ is injective, tensorizing by -$\otimes_AB$ we get an injective ring homomorphism $$\phi\colon B/\mathfrak qB=B\otimes_A A/\mathfrak q\longrightarrow B\otimes_AB/\mathfrak p.$$ Now just notice that $\phi$ factorizes as $$B/\mathfrak qB\longrightarrow 
B/\mathfrak p \longrightarrow B\otimes_AB/\mathfrak p.$$
I know it has been a while since the question was posted, anyway (if I'm right) the following should be an example of $\mathfrak p^{ce}=\mathfrak p.$

Let $f\colon A\longrightarrow B$ be an injective, flat ring homomorphism. Let $\mathfrak p\in\mathsf{Spec}(B)$ and $\mathfrak q:=f^{-1}(\mathfrak p)$. I claim that $$\mathfrak qB=\mathfrak p.$$

The ring map $B/\mathfrak qB\longrightarrow B/\mathfrak p$ is onto. We show that it is injective too. As $f$ is flat and $A/\mathfrak q\longrightarrow B/\mathfrak p$ is injective, tensorizing by -$\otimes_AB$ we get an injective ring homomorphism $$\phi\colon B/\mathfrak qB=B\otimes_A A/\mathfrak q\longrightarrow B\otimes_AB/\mathfrak p.$$ Now just notice that $\phi$ factorizes as $$B/\mathfrak qB\longrightarrow 
B/\mathfrak p \longrightarrow B\otimes_AB/\mathfrak p.$$