 Solution

Write $C_n=⟨g⟩$ and set $N=⟨g^p⟩$. The order of $g^p$ is $n/\gcd (n,p)=m$, so $N\cong C_m$. Every subgroup of an abelian group is a normal subgroup. The quotient group $C_n/N$ is generated by $gN$ and has order $n/m=p$, hence is $C_p$ and is a simple group. By the subgroup correspondence theorem, there is no proper normal subgroup strictly between $N$ and $C_n$. Therefore $N$ is a maximal normal subgroup with the required quotient group.