$\mathcal{O}/(\mathcal{O}\cap I\mathcal{O}_P)\cong \mathcal{O}_P/I\mathcal{O}_p$ if $\mathcal{O}$ is a noetherian one dimensional domain. 
Let $\mathcal{O}$ be a Noetherian, one dimensional domain and $P\subset \mathcal{O}$ be a prime ideal. I want to prove that if $0\neq I\subseteq P$ we have $\mathcal{O}/(\mathcal{O}\cap I\mathcal{O}_P) \cong \mathcal{O}_P/ I\mathcal{O}_P$.

It's clearly enough to prove that the map $\mathcal{O}\rightarrow \mathcal{O}_P/ I\mathcal{O}_P$ is surjective, but I can't prove that.
This is used in the proof of propostion 12.3 in Neukirch Algebraic Number Theory.
 A: Since $\mathcal{O}$ is a $1$-dimensional domain and $I\neq 0$, $\mathcal{O}/I$ is $0$-dimensional.  Since it is Noetherian, that means $\mathcal{O}/I$ is the product of its localizations at each of its prime ideals.  In particular, the map from $\mathcal{O}/I$ to any localization of $\mathcal{O}/I$ is surjective (since it is just the projection onto a factor of a product).  Thus the map $\mathcal{O}/I\to \mathcal{O}_P/I\mathcal{O}_P$ is surjective, which gives the isomorphism you want.
A: Here is the argument that I suspect Neukirch intended.
In the first part of the proof of Prop 12.3 in Neukirch's book, he states that
if $P \supseteq I$, then $P$ is the only prime ideal containing $\widetilde{I}_P$
where $\widetilde{I}_P = (\mathcal{O} \cap I\mathcal{O}_P)$.
He doesn't offer a proof, but a simple one is given in Proof of a stronger form of Chinese remainder theorem (12.3) in Neukirch
That means that $\mathcal{O}/\widetilde{I}_P$ has a single maximal ideal, $P/\widetilde{I}_P$ and so
everything not in that ideal is a unit.  Therefore, if $x/s \in \mathcal{O}_P$ with
$s \notin P$, and writing modulo $\widetilde{I}_P$ with an overbar, $\overline{s}$
has an inverse $\overline{s}' \in \mathcal{O}/\widetilde{I}_P$ so that
$xs' \in \mathcal{O}$ maps to $\overline{x}/\overline{s}$, showing surjection.
