for the table R(A,B,C,D), the candidate keys are AB and BD respectively.
Now this table is already in third normal form but not in BCNF due to the functional dependency D->A.
So we find out the closure of D using the functional dependencies applicable on R and it contains {D,A}.
So we split the table R into two tables: R1(BCD) and R2(AD).
Here, decomposition is lossless. Only the FD D->A is applicable on R2 and the only FD that is applicable for table R! is BD->C since BD is the candidate key. Using these two functional dependencies, there is no way in which we can derive the FD AB->CD. Hence this FD is not applicable on any of the decomposed tables so the BCNF decomposition is not dependency preserving.