I don't understand why $\mathfrak b=\mathfrak a_{\mathfrak p}$. Since $\mathfrak b$ is a fractional ideal of $A$, we have that $\mathfrak b\subset Q(A)$ (where $Q(A)$ is the quotient field of $A$), but if we contract $\mathfrak b$ to $\mathfrak b\cap A$, we get an ideal in $A$. Its extension $\mathfrak a_{\mathfrak p}$ can't be equal to $\mathfrak b$ (if $\mathfrak b$ wasn't an ideal of $A$ to begin with), because the extension is going to be an ideal in $A$ (and not just a fractional ideal).
Where am I making a mistake?
Note that $\mathfrak a_\mathfrak p=\{a/s:a\in\mathfrak a,s\notin\mathfrak p\}$. I will show both inclusions.
Why $\mathfrak b\subseteq\mathfrak a_\mathfrak p$:
Let $a/s\in\mathfrak b$ where $a\in A$ and $s\in A\setminus\mathfrak p$ be arbitrary. Then clearly $a=s\cdot a/s\in\mathfrak b\cap A=\mathfrak a$, so that $a/s\in\mathfrak a_\mathfrak p$.
Why $\mathfrak b\supseteq\mathfrak a_\mathfrak p$:
Let $a/s\in\mathfrak a_\mathfrak p$ be arbitrary. Here $a\in\mathfrak a\subseteq\mathfrak b$, so $a/s=1/s\cdot a\subseteq \mathfrak b$.