HINT: $a^k\bar a^k=(a\bar a)^k$
Added: (That was written before you edited the question.) For any integer $k$ you have $a^k\bar a^k=(a\bar a)^k=1^k=1\;.\tag{1}$ Take $k=\operatorname{ord}_m(a)$: $(1)$ becomes $\bar a^{\operatorname{ord}_ma}=a^{\operatorname{ord}_ma}\bar a^{\operatorname{ord}_ma}=1\;;\tag{2}$ take $k=\operatorname{ord}_m\bar a$ instead, and it becomes $a^{\operatorname{ord}_m\bar a}=a^{\operatorname{ord}_m\bar a}\bar a^{\operatorname{ord}_m\bar a}=1\;.\tag{3}$
I expect that you know that if $a^n=1$, then $\operatorname{ord}_ma\mid n$, i.e., $n$ is a multiple of $\operatorname{ord}_ma$. (If not, that’s the first thing that you need to prove.) If you combine that fact with $(2)$ and $(3)$, it’s not hard to prove that $\operatorname{ord}_ma=\operatorname{ord}_m\bar a$.