If \(S\) and \(T\) have different number of W and B, then the answer is No. Now we assume that they are equal. Let \(W\) be the number of W and \(B\) of B.

Since the empty squares are always adjacent to each other, there are \(\frac{(N+1)!}{B!W!}=O(\sqrt{N}2^N)\) possible states of the cells. (The analysis is a bit complicated, so we omit it here.)
This takes on the maximum value \(51480\) for \(N=14\) and \(B=W=7\). For each state, there are \((N-1)\) possible moves.

Since the number of states and the next moves are both small enough, the problem can be solved with BFS.

The total complexity is (expected) \(O(N^{1.5}2^N)\) (using a hash map).

Similar problem: 8 puzzle (AOJ)