The number of cipher texts possible for each plaintext This is from Dan Boneh's book
Theorem 2.1. Let X = (E, D) be a Shannon cipher defined over (K, M, C). The following are
equivalent:
(i) X is perfectly secure.
(ii) For every $c \in C$, there exists $N_c$ (possibly depending on c) such that for all $m \in M$, we have
$|\{k \in K : E(k, m) = c\}| = N_c$
(iii) If the random variable k is uniformly distributed over K, then each of the random variables E(k, m), for $m \in M$, has the same distribution.
Proof:
For every $c \in C$, there exists a number $P_c$ (depending on c) such that for all $m \in M$, we have Pr[E(k, m) = c] = $P_c$. Here, k is a random variable uniformly distributed over K. Note that $P_c = N_c/|K|, where N_c$ is as in the original statement of (ii)
(Partially copied, not the full thing)

Point (ii) is not clear to me. What exactly is $N_c$? If is the number of ciphertexts possible for each plaintext, then it's always equal to 0 or 1 for perfect secrecy, right. Or can it be something else?
What exactly is $N_c$?

It is a number that may depend on a given ciphertext and states how many keys exist for every message that produce this ciphertext. The statement here is that every ciphertext can be produced from every message with an equal number of keys, i.e. if you draw keys uniformly at random all messages had the same probability of being the source.
For perfectly secret schemes with $|K|=|M|$ this will always be either 1 or 0. It could be 0 e.g. if the ciphertext is longer than anything in the message space for a OTP.