B -> AD will fail 2NF
B -> A (prime gives prime) and B -> D (prime gives non prime)
To pass 3NF, LHS of FD should be either super key or RHS should be a prime attribute for ALL FD
But for B -> D neither holds so it isn't in 3NF
To be in 2NF, RHS of FD should be given by an any of the complete Candidate Key not a subset of it. (No partial dependency allowed)
B -> D (subset of candidate key gives D )
So it's not even in 2NF.
Here,