Fermat's Little Theorem :
$$a^p ≡ a (\mod p)$$
According to Modular Arithmetic $a ≡ b (\mod n)$ if their difference $(a-b)$ is an integral multiple of $n$ $( n$ divides $(a-b) )$
So, $( a^p - a )$ is an integral multiple of $p.$ Now as $a$ is not divisible by $p$ so definitely, $( a^{p-1} -1)$ is an integral multiple of $p.$ This simply means if we divide $a^{p-1}$ by $p,$ the remainder would be $1$. i.e., $ a^{p-1} \mod p = 1.$
Put the values in the formula.
$p=17$
So, $p-1 =16.$
Correct Answer: $D$