Clearly, there cannot be more than $p$ different numbers. It follows from the condition of the problem, that $0$ could not be someone's number on the first step, since then that person will have it on the second step as well. Also, no one can have $1$ as a first number, since then the neighbor on the left will have two equal numbers. We will show that the conditions hold for $N = p-2$.

Indeed, let the following consecutive numbers be written down on the first step from left to right: $2, 3, 4, \ldots, p-1$. They are all pairwise distinct. Assume that after the second step someone's two numbers are equal. If it's one of the first $N-1$ people, it means that the product $k(k+1)$, where $2 \le k \le p-2$, has the same remainder when divided by $p$, as the number $k$. But then their difference is divisible by $p$, that is, $k(k+1)-k=k^2$ is divisible by a prime number $p$, thus, $k \nmid p$, which is not possible. If the $N$-th person has two equal numbers, then the number $p-1$ will be multiplied by $2$, but then two numbers will be different, since $2(p-1)-(p-1)=p-1$ is not divisible by $p$.