How many elements of order 7 in a simple group of order 168?

The $8$ Sylow subgroups of order $7$ each contain $6$ elements of order $7$ together with the identity. Since an element of order $7$ generates a group of order $7$ these elements are all distinct. There are $8 \times 6=48$ elements of order $7$.

You use the fact that the group is simple in asserting that there are $8$ subgroups of order $7$. It is always possible to have $1$ subgroup of a given order (so long as it is a factor of the group order e.g. the cyclic group of order $168$) - but a single subgroup of order $7$ would be normal, and the group would not be simple.

Making use of the theorem-

Let $G$ be a group of order $p^{\alpha}m$,where $p$ is a prime,$m\ge 1$ and $p$ does not divide $m$,then the number of sylow $p-$subgroup of $G$ is of the form $1+kp$ i.e,$$n_p \equiv 1(modp)$$.Further,$n_p$ is the index in $G$ of the normalizer $N_G(P)$ for any Sylow $p-$ subgroup of $P$,hence,$n_p$ divides $m$.

Since,$\vert G\vert=168=2^3\cdot 3\cdot 7$;

$n_7=1+7k$ for some integer $k$,and $n_7$ divides $2^3\cdot 3=24$,this restricts the only possibility for $k$ to be $1$ only. So, $n_7=1+7\cdot 1=8$.