Answer is 2 because
Consider a relation R(ABCDE) with fds (AB->CDE, C->A, D->E)
Primary keys are AB, BC (by closure properties) .
B alone can't be candidate key and therefore C is not superkey.
Now try to decompose the table with the definition of BCNF
X->y
(X is a superkey only )
Now when decomposing C will take attribute A with itself coz (C->A) and form a new table with itself, thus preserving C->A dependency.
Another table will be BCDE (with primary key BC)
this table will be decomposed into 2 tables DE and BCD
DE will preserve D->E and BCD is table as it contains primary key for joining operation.
Now our tables are
CA, DE, BCD
all these tables are in BCNF but we see that our dependency AB->CDE is lost , therefore it is not fd preserving hence answer is option 2.
