For this, first we select two letters whose position is correct.. This can be done in 4C2 ways = 6 ways..
For the remaining two letters, we need to do derangement which is done in D2 = 2!/2 = 1 way.. (the only way to derange two objects is to swap them)..
Hence number of ways = 6 * 1 = 6.. Hence D) should be correct option..