There can be more than one valid BCNF decomposition depending on which violation is selected first.
Take the violation $C\to AD$.
Compute $C^+$ in the original relation.
Start with $C$.
Using $C\to AD$,
obtain $A$ and $D$.
Using $D\to B$, obtain $B$.
Thus,
$C^+=\{A,B,C,D\}$
$E$ is missing, so $C$ is not a superkey of $R$.
Therefore $C\to AD$ violates BCNF.
Using the closure-based split gives
$R_1(A,B,C,D)$ and $R_2(C,E)$.
$R_2(CE)$ has no non-trivial projected FD, so it is in BCNF.
Now inspect $R_1(ABCD)$.
The dependency $D\to B$ still holds.
But $D^+=\{D,B\}$
within $R_1$, so $D$ is not a superkey.
Decompose $R_1$ into
$R_{11}(D,B)$ and $R_{12}(A,C,D)$.
In $DB$, $D$ is a key.
In $ACD$, the projected dependency
$C\to AD$ makes $C$ a key.
Hence both are in BCNF.
The final decomposition is $\boxed{{DB,\ ACD,\ CE}}$
Answer : C
