This assumption doesn’t hold in general, however you can construct an efficient algorithm, even if it doesn’t hold.

First, let’s show that a cycle always exists. Let I be the size of the instruction string, and N be the number of nodes. Since the number of states for each ghost is at most I*N, after a finite number of steps, the ghost will go into one of the previous states and cycle forever. Let’s say that the cycle length is c, and after a+c steps the ghost has entered the same state it was after a steps.

Let’s assume[^1] that during the first a+c steps the ghost has only once encounter an end state (a node ending with ‘Z’), specifically after e states. If e >= a, this means that the ghost will encounter the end state also after e + c and e + 2c and so on, or for every number of steps s > e such that s = e (mod c). The assumption you formulated means e = 0 (mod c), or e = c.

Now, consider the K ghosts that are travelling simultaneously. If after n steps all ghosts have reached the end state, this means that n = e_i (mod c_i) for all ghosts i (1 <= i <= K). According to the Chinese remainder theorem, there is a solution if and only if e_1 = e_2 = ... = e_K (mod gcd(c_1, c_2, ... c_K)). If the assumption you formulated holds, then e_i = 0 (mod c_i), so lcm(c_1, ... c_K) works as a solution. If it doesn’t, you can still find n, but it will be a bit more tricky (which is probably why the authors of the challenge made e = c always).

[^1] – this is another assumption you’ve implicitly made, and that happens to hold for all the inputs. However, if this assumption doesn’t hold, we can check all possible combination of end state positions.