Any problem in P is also in NP
A decision problem that's in P is also in NP, because you can give the verification logic like this: for yes instance
x, use empty string as a certificate, and solve
x in polynomial time. You get the result that it's yes instance (that's by definition of P) and that means verification is done in polynomial time.
Note that, the order written as $n^2 k^2$ doesn't simply mean the problem is in P (Check "Pseudo-polynomial time" in wikipedia.) For example, Knapsack problem can be solved by the dynamic programming and its order is $O(n W)$ where $n$ is the number of products and $W$ is the maximum weight for the knapsack. But $W$ is actually considered exponential of the input size and Knapsack problem is known to be NP-complete.