Let $km$ be the order of $a$ modulo $n$. Show that $k \mid \phi (n)$.
Since $km$ is the order of $a$ modulo $n$, we have $\gcd(a,n)=1$, and so Euler's theorem tells us
$$ a^{\phi(n)} \equiv 1 \pmod n \tag{i}$$
By definition of order, we have
$$ a^{km} \equiv 1 \pmod n \tag{ii}$$
where $km$ is the smallest positive $j$ such that $a^j \equiv 1 \pmod n$.
Comparing (i) and (ii), this means $km$ divides $\phi(n)$, and so $k$ also divides $\phi(n)$. That is
$$ k \mid \phi(n) $$