Extension of a discrete non-archimedean absolute value

Let $P$ be a prime ideal of a Dedekind domain $A$, $v$ an absolute value induced by $P$, and $B$ the integral closure of $A$ in a finite separable extension $E$ of $A$'s quotient field $K$. If $w$ is an extension of $v$ to $E$, what exactly is the valuation ring $B_w = \{ x \in E : |x| \leq 1\}$? Is it $B_{\mathfrak B}$ for some prime ideal $\mathfrak B$ of $B$ lying over $P$?

Of course $w$ is still non-archimedean, since this is equivalent to saying $|y| \leq 1$ for all $y$ in the additive group generated by $1$. It follows that $B_P \subseteq B_w$; $B_P$ is the integral closure of $A_P = \{ x \in K : |x| \leq 1\}$, and if $x \in L$ is integral over $A_P$, then $0 = a_0+a_1x+ \cdots + x^n$ for some $a_i \in A_P$. If $|x|$ is not $\leq 1$, then $$|x|^n \leq Max\{|a_0|, |a_1x|, ... , |a_{n-1}x^{n-1}| \} \leq Max\{1, |x|, ... , |x|^{n-1}\} \leq |x|^{n-1} < |x|^n$$

I think the answer to your question is yes. Let $\mathfrak B = \{x \in B : |x| < 1\}$. This is a prime ideal of $B$, and nonzero because $E$ is the quotient field of $B$. As you said, $B_P \subseteq B_w$, so $|x| \leq 1$ for all $x \in B$.
It follows that $B_{\mathfrak B} = B_w$. If $a/s \in B_{\mathfrak B}$, we can assume $a \in B$ (hence $|a| \leq 1$) and $s \in B \setminus \mathfrak B$ (so $|s| = 1$), and so $|a/s| \leq 1$.
Conversely if $x \not\in B_{\mathfrak B}$, the fact that $E$ is the quotient field of $B_{\mathfrak B}$ means that $1/x \in B_{\mathfrak B}$, so we can write $1/x = a/s$ with $s \in B \setminus \mathfrak B$ and $a \in \mathfrak B$ (if $a \in B \setminus \mathfrak B$ then $x = (a/s)^{-1} \in B_{\mathfrak B}$, which we are supposing to not be true). Then $|1/x| = |a||s|^{-1} < 1$, so $|x| > 1$ and hence $x \not\in B_w$.