I want to prove the statement in the title, but I think we need $d \geq 2$ in the statement since otherwise there is a case not fulfilling the statement. My attempt:
By assumption we have $ord_p(a)=:d$, which means that $d$ is the least natural number such that $a^d \equiv 1 \mod p$. Then
$$a^d -1 = (a-1)(a^{d-1}+\dots+a^0) \equiv 0 \mod p$$
Since $\Bbb Z_p$ is a field, there are no zero divisors in $\Bbb Z_p$. It follows that $a-1 \equiv 0 \mod p$ or $a^{d-1}+\dots+a^0 \equiv 0 \mod p$. Now I would need $ord_p(a) \neq 1$, i.e. $a \not \equiv 1 \mod p$.
Otherwise we might have $a-1 \equiv 0 \mod p$, then $a \equiv 1 \mod p$, which would mean $d=ord_p(a)=1$. But then $a^{d-1}+\dots+a^0=1 \equiv 1 \mod p \not \equiv 0 \mod p$.
So I would get that if $a=1$ in $\Bbb Z_p$, an thus $ord_p(a)=1$, the statement is not true. So it should say $d \geq 2$ in the statement, shouldn't it? Or am I missing something?