Since this is homework, I'm not going to give the answer, but will hopefully point you in the right direction.
You are correct when you say for a perfect cipher, the probability should hold that $\mathbb{P}(P=m|C=c)=\mathbb{P}(P=m)$. In words, this means that given that you see the ciphertext is $c$ what is the probability of the plaintext being $m$ (note, I'm using $P$ for the plaintext space and $C$ for the ciphertext space). If being given the ciphertext does not help the adversary at all (i.e., the probability of any plaintext does not change at all given the ciphertext), then the cipher is said to be perfect. Almost all modern ciphers that we use today we cannot prove this property. For the cipher you are studying, however, it is possible.
By applying bayes rule, we get
$\mathbb{P}(P=m|C=c)=\frac{\mathbb{P}(C=c|P=m)\mathbb{P}(P=m)}{\mathbb{P}(C=c)}$ and we want to show that this equals $\mathbb{P}(P=m)$. Therefore, if the other term on top cancels out the term on bottom, we would get what we want.
To help a little more, think about what the two terms are. On bottom we have $\mathbb{P}(C=c)$. In other words, what is the probability of each ciphertext (independent of plaintext). The ciphertext space is $\mathbb{Z}_p$ so the size of it is $p$.
The other term is $\mathbb{P}(C=c|P=m)$, meaning, given a plaintext message $m$, what is the probability that the ciphertext is some $c$. So, given a plaintext $m$ if all ciphertexts in $C$ are produced by the same number of keys, then $\mathbb{P}(C=c|P=m)$ would simply be $\mathbb{P}(C=c)$ in which case the two terms would cancel as we want.
I'll leave that up to you to show.
