I have to perform the computation of $a^{b}\pmod {m}$. However, instead of providing $b$ directly, I instead know the value of $b\mod {m}$. Is it even possible to perform this calculation and if so, how do I go about it?
EDIT: I know that $m$ is a prime number.
As stated, this is not possible. Taking $m=3,\;a=2$ we remark that $$2^1\not \equiv 2^4\pmod 3$$
What you need is $b\pmod {m-1}$:
Assume that $a\not \equiv 0\pmod m$. Then $a^{m-1}\equiv 1 \pmod m$, by Little Fermat. Thus if you replace $b$ with $b+k(m-1)$ we get $a^{b+k(m-1)}=a^b\times \left(a^{m-1}\right)^k\equiv a^b\pmod m$.