To get you started some rather basic (or even trivial, depends) observations, putting $\,n_p=$ number of Sylow $p-$subgroups , $\,G:=\,$ simple group of order $\,168\,$:
$$168=2^3\cdot 3\cdot7\Longrightarrow n_7=8\;,\;n_3=7,28$$
It can't be $\,n_3=4\,$ as then $\,G\,$ has a subgroup of index 4, which is impossible since this would mean $\,G\,$ is isomorphic with a subgroup of $\,S_4\,$ . For the same reason, it must be $\,n_2=7,\,21\,$...
Perhaps reading here you'll have have the whole view. Don't worry about the number of pages as the first ones are basic results listed.
You can see that if G is noncyclic abelian then $G$ be must be $Z_{10}\times Z_2$.
Claim: $H=Z_5\times1$ is uniqe subgroup of $G$ with order $5$.
if there is also $K$, $HK$ has $25$ elements which is impossible.
Now,Any subgroup of order $10$ must include $H$ and $G/H\cong Z_2\times Z_2$ Since $G/H$ has three subgroup of index $2$, $G$ has three subgroup of order $10$.
Best Answer
First, suppose $p<q$. Then $n_q\equiv 1\pmod{q}$, and $n_q\mid p^2$. So the only possibilities are that $n_q=1,p,p^2$. If $n_q=1$, the group is not simple. But $n_q=p$ is impossible, else $q\mid p-1$. So $n_q=p^2$. Then we have $p^2$ groups of prime order $q$, so they have pairwise trivial intersection, and this forces the $p$-Sylow subgroup to be unique.
The $q$-Sylow subgroups give a total of $p^2(q-1)$ elements of order $q$. This accounts for all but $p^2$ elements of the group, so any $p$-Sylow subgroup must be these remaining elements, hence unique.
The case where $q<p$ is quite quick. Any $p$-Sylow subgroup has index $q$. This is smallest prime dividing $|G|$, so your $p$-sylow subgroup is normal.