Like in my previous question (Exercise about primary ideals), consider the canonical homomorphism $A\to A_\mathfrak p$, for a fixed prime ideal $\mathfrak p$ in the ring $A$, and write $I\cap A$ for the contraction of any ideal $I\subseteq A_\mathfrak p$. Show that if a $\mathfrak p $-primary ideal contains $\mathfrak p^n$, then it contains $\mathfrak p ^{(n)} := \mathfrak p ^nA_\mathfrak p\cap A$.
Take $\mathfrak q\supseteq \mathfrak p^n$ a $\mathfrak p$-primary ideal: if $a\in \mathfrak p^{(n)}$, exists a $b\in A\setminus \mathfrak p$ such that $ab\in \mathfrak p^n$, i.e. $ab\in \mathfrak q$ and $b\notin \mathfrak q$. However this proves that $a\in \sqrt {\mathfrak q}=\mathfrak p$, not that $a\in \mathfrak q$. Can you just give me an hint?
Let $\mathfrak q\supseteq \mathfrak p^n$ be a $\mathfrak p$-primary ideal and $a\in \mathfrak p^{(n)}$. There exists $s\in A\setminus \mathfrak p$ such that $sa\in \mathfrak p^n$, hence $sa\in \mathfrak q$ and $s\notin \mathfrak p$. It follows that $a\in \mathfrak q$.