Show that $I=(X_{k_1}^{a_1},...,X_{k_s}^{a_s})$ is $(X_{k_1},...,X_{k_s})$-primary.
I noticed that $$\sqrt{({X_{i_1}}^{a_1},...,{X_{i_k}}^{a_k})}=\sqrt{({X_{i_1}}^{a_1})+\cdots+({X_{i_k}}^{a_k})}=\sqrt{\sqrt{(X_{i_1})^{a_1}}+\cdots+\sqrt{(X_{i_k})^{a_k}}}=\sqrt{(X_{i_1})+\cdots+(X_{i_k})}=\sqrt{(X_{i_1},...,X_{i_k})}=(X_{i_1},...,X_{i_k}).$$ I have to show that $I$ is $P$-primary (where $P=(X_{k_1},...,X_{k_s})$), i.e., if $u v\in I$ then $u\in I$ or $v\in P$. I tried to prove that if $u\notin I$ and $v\notin P$ then $uv\notin I$. (There is a similar question, but the answer is too hard for me.)
If $u\notin I$ and $v\notin P$ then there is a monomial $m$ from $u$ such that $m\notin I$, and a monomial $n$ from $v$ such that $n\notin P$ (see here). But $mn\in I$, so there is $t$ such that $X_{k_t}^{a_t}\mid mn$. Can you get a contradiction?