Proof :
We can assume that p$ \neq $ q and $n_p >1 $, so $n_p$ = q. Now choose distinct Sylow p- subgroups $S$ and $T$ of $G$ such that $|S\cap T|$ is as larger as possibe and write $D = S \cap T$.
if $D$ = 1 , then every pair od distinct Sylow p - subgroups of $G$ intersect trrivally, So the total number of elements whose order is of the form $p^b$ is $q(p^a -1)$ and the number of element is atmost $|G| -q(p^a - 1) = q$ elements with order not divisble by p. itfollows that if $Q \in Syl_q(G)$, then $Q$ is exactle the set of these elements and $Q$ is unique and $Q$ is normal in $G$, $G$ is not simple, as required.
I want to help to prove that if $|D|$ >1