I'm trying to write a different proof from the book I'm using since I am seeing a connection to this corollary from the same book. However I am not certain if the connection I see is correct.
Corollary: Let $a$ have order $k$ modulo n, then $a^h$ has order k if and only if $gcd(h,k)=1$.
Proof of the title of this post:
Suppose $d|(p-1)$ and $a^d\equiv 1\pmod p$ such that order of $a$ is $d$ modulo p. Thus $\forall i, 1\leq i\leq d-1$, $a^i \not\equiv 1\pmod p$. By the corollary, There exists $\phi(d)$ $a^i$'s such that the order of $a^i$ modulo p = $d$. Since every $a^i \equiv k\pmod p$ such that $1\leq k\leq p-1$. Then by a bijective relation between all the powers of $a$ and $k$, there exist $\phi(d)$ integers between and including $1$ and $p-1$ having order $d$ modulo p.