How do i see easily that $B'/\mathfrak{p}B'=B/\mathfrak{p}B$?

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I'm reading a book "Local Fields", J.Serre. There is something I don't understand. First, I will explain the situation.

$A$ is a Dedekind domain, $\mathfrak{p}$ is a nonzero prime ideal of $A$, $\mathbf{Frac}\{A\}=K$, $L$ is a finite separable extension of $K$ and $B$ is a integral closure of $A$ in $L$.

Now, in 15p, Serre said

Let $S=A-\mathfrak{p}$, $A'=S^{-1}A$, and $B'=S^{-1}B$. The ring $A'=A_{\mathfrak{p}}$ is a discrete value ring, and $B'$ is its integral closure in $L$. One has $A'/\mathfrak{p}A'=A/\mathfrak{p}$, and one sees easily that $B'/\mathfrak{p}B'=B/\mathfrak{p}B$.

I can prove that $A'/\mathfrak{p}A'=A/\mathfrak{p}$. But i don't know how to see easily $B'/\mathfrak{p}B'=B/\mathfrak{p}B$. It is much harder to me since $\mathfrak{p}B$ is not a prime ideal of $B$.

My attempt : Consider the canonical map from $B/\mathfrak{p}B$ to $B'/\mathfrak{p}B'$. This ring-homomorphism is well-defined. But is it injective? (namely, $\mathfrak{p}B'\cap B=\mathfrak{p}B$ ?),, is it surjective?

Please, help me.