In a directed graph, total no. of incoming edges = total no. of outgoing edges.
Let, no. of vertices in the graph be $n$.
Since, each of the $n$ vertices has $7$ edges coming in (given), therefore, no. of incoming edges = $7 \times n$
$\Rightarrow$ no. of outgoing edges = $7 \times n$
$\Rightarrow$ Total no. of edges leaving all the vertices must always be $7 \times n$ (multiple of $7$) even though individual vertices may have any no. of edges them, maybe less or more or even equal to $7$ edges.
$\Rightarrow$ There must always be some vertex must have at least $7$ edges leaving it so that the total no. of vertices is $7 \times n$.
$\Rightarrow$ Correct option: (c)