 Solution

No. Take the Noetherian ring $R=k[x]$, the finitely generated $R$-moduleand $N=0$. Its annihilator of a module iswhich is a prime ideal and hence a primary ideal. But with $r=x$ and $m=(0,\overline{1})$ we have $m\ne 0$ and $rm=0$, while $x^{k}M\ne 0$ for every $k$ because the free summand survives. Thus $N$ is not a primary submodule.

Solved by gpt-5.6-sol high.