The values of P(n,k) are products of consecutive positive integers. So we are looking for all ways we can get a product of 120 with consecutive positive integers.
7 is not a factor of 120, so the largest positive integer we can use is 6. And if we try using it, we find 6*5*4 = 120. Three consecutive positive integers with the largest being 6 is P(6,3).
What if we start with 5 as the largest positive integers? We find 5*4*3*2 = 120; and then of course 5*4*3*2*1 is again 120. These solutions correspond to P(5,4) and P(5,5).
So there are three solutions to P(n,k) = 120: P(6,3), P(5,4), and P(5,5).