This puzzle is not solvable. The proof involves noting that there are two distinct sets of positions which can be assembled from the pieces with different parity, and there is no way of moving between them using the allowed moves, as they preserve parity. The parity in this context (the invariant) is the parity (odd or even) of the number of pairs of pieces in reverse order plus the row number of the empty square. For the order of the 15 pieces consider line 2 after line 1, etc., like words on a page.

Thus an even permutation of the order of the 15 pieces can only be obtained if the empty square is not moved or moved two rows, and an odd permutation of the order of the 15 pieces can only be obtained if the empty square is moved one or three rows.

By contrast, note that considering the parity of permutations of all 16 squares (15 pieces plus empty square) is not meaningful here, because it changes with every move.