 Solution

Reducing coefficients modulo $p$ gives the ring isomorphismIf $d=\mathrm{deg}f>0$, its monic leading coefficient remains one modulo $p$, so $\mathrm{deg}\bar{f}=d$. Polynomial division gives a unique representative of degree less than $d$: existence follows by division, and the difference of two such representatives cannot be a nonzero multiple of a degree-$d$ monic polynomial. There are $p$ choices for each of the $d$ coefficients, provingIrreducibility of $\bar{f}$ is not needed for finiteness; without it the quotient ring need not be a field.