$\DeclareMathOperator{\Ann}{Ann}\DeclareMathOperator{\Rad}{Rad}$Let $R$ be a noetherian, reduced ring which is not an integral domain (and may also not be factorial).
Let $P$ be a minimal prime of $R$. Is it possible to find $b \in R$ with $P = \Ann(b)$ and $bR$ radical?
What I tried: Consider a non-zero element $d \in \Rad(bR)$. By assumption we have $\Ann(d) \subseteq \Ann(b) = P$ since $P$ is prime and $d \neq 0$. Moreover, since $R$ is reduced, we obtain equality, i.e. $\Ann(d) = \Ann(b) = P$. But I don't know how to proceed from here on.
Note that we may change $b \in R$ in any way that preserves $\Ann(b) = P$. We may also assume (if necessary) that $R$ is finite free over some principal ideal domain (since the original motivation is from geometry where I consider a projective curve lying finitely over the projective line).
I am grateful for any kind of help.
Edit: I edited the question since it wasn't clear what I meant.
This question is related to the following: Is $R/bR$ reduced if $R$ is reduced and $\operatorname{Ann}(b)$ is a minimal prime of $R$?