Given $R(A,B,C,D,E)$ with $\Sigma=\{A\to B,\ AE\to D,\ B\to E,\ AD\to E,\ CD\to A,\ AB\to D,\ E\to B\}$.
From $A\to B$ and $B\to E$,
we get $A\to E$.
Also, since $A\to B$ and $AB\to D$,
we get $A\to D$.
Hence, $A\to BDE$.
Now $CD\to A$, and since $A\to BDE$, $CD\to ABCDE$.
So $CD$ is a candidate key of $R$.
Consider the decomposition $\{ABD,\ ACD,\ BE\}$.
In $ABD$, $A\to B$ and $A\to D$,
so $A$ is a key. Hence $ABD$ is in BCNF.
In $BE$, $B\to E$ and $E\to B$, so both $B$ and $E$ are keys.
Hence $BE$ is in BCNF.
In $ACD$, $CD\to A$ and $A\to D$.
Here $A\to D$ violates BCNF because $A$ is not a superkey of $ACD$.
However, $AC$ and $CD$ are candidate keys of $ACD$, so $D$ is a prime attribute.
$\therefore A\to D$ satisfies $3$NF.
Hence $ACD$ is in $3$NF but not BCNF.
The decomposition is dependency-preserving because
$A\to B$ and $A\to D$ are preserved in $ABD$,
$B\to E$ and $E\to B$ are preserved in $BE$,
and
$CD\to A$ is preserved in $ACD$.
The remaining dependencies follow from these.
For example,
$AE\to D$ follows from $A\to D$,
$AD\to E$ follows from $A\to B$ and $B\to E$,
and
$AB\to D$ follows from $A\to D$.
The decomposition is also lossless because the synthesis contains a candidate key, namely $CD$, inside $ACD$.
Answer : A
