Understanding the third Sylow theorem I am trying to understand the following theorem:
Let $G$ be a finite group, $p$ a prime number, and let's suppose $|G|=p^ns$ s.t. $p$ doesn't divide $s$
Let $n_p$ be the number p-sylow subgroups of  G then we have

$$\begin{cases}n_p |s\\n_p  \equiv 1 \mod p\end{cases}$$

Now my problem is with $n_p$ I don't understand why $n_p$ can be anything different from 1. a p-sylow subgroup is a p-subgroup which is maximal, but I don't understand why there can be more than one maximal subgroup of a group G

Maximal means "not contained in anything else" not "everything else is contained in it."
Look at $S_3$ where you have three Sylow-$2$ subgroups, $\{(1), (ij)\}$ for any $1\le i\ne j\le 3$.
Or if you prefer a simpler example of where "maximal" means this, look at a case with some sets: consider $\{\varnothing, \{1\},\{2\}\}$ with respect to inclusion both $\{1\}$ and $\{2\}$ are maximal because they are not properly contained in anything larger, even though neither contains the other.
