It is given that the relation $R(A, B, C, D)$ is in BCNF. And we know that one out of the given four fds does not hold. i.e. only 3 holds.
(A) $A \rightarrow BCD$
(B) $BC \rightarrow A$
(C) $CD \rightarrow B$
(D) $D \rightarrow C$
Let's start by option A, i.e. we assume that A does not hold, while B, C and D holds
So, the candidate key will be: $D$, since $D^+ = \{ A, B, C, D\} $
But the fd $BC \rightarrow D$ does not satisfy the BCNF requirement. So, the $R(A, B, C, D)$ with three fds (B), (C) and (D) is not in BCNF.
Looking at option B, i.e. except B all three others are fds that hold.
We have candidate key $A$. Here $D\rightarrow C$ violates BCNF.
Looking at option C.
candidate keys are: $A$, $BC$ and $BD$.
here, $D\rightarrow C$ violates BCNF.
We are left with last option D.
Candidate key will be $A$, $BC$, $CD$. And these satisfy the BCNF requirement.
So, we can conclude that fd of option (D) is the one that should not hold for $R(A, B, C, D)$ to be in BCNF.