Here, we can see that $D$ is not part of any $FD's$, hence $D$ must be part of the candidate key.

Now $D^{+}$=$\left \{ D \right \}$.

Hence, we have to add A$,B,C,E,F,G,H$ to $D$ and check which of them are Candidate keys of size $2$.

We can proceed as:

**AD+= {A,B,C,D,E,F,G,H}**

Similarly we see **BD+** , **ED+** and **FD+** also gives us all the attributes. Hence, **AD,BD,ED,FD** are definitely the candidate keys.

But CD+, GD+ and HD+ doesnnt give all the attributes hence, $CD,GD$ and $HD$ are not candidate keys.

Now we need to check the candidate keys of size $3$. Since $AD, BD, ED, FD$ are all candidate keys hence we can't find candidate keys by adding elements to them as they will give us superkeys as they are already minimal. Hence, we have to proceed with $CD$, $GD$ and $HD.$

Also, we can't add any of ${A,B,E,F}$ to $CD$, $GD$, $HD$ as they will again give us superset of ${AD,BD,ED,FD}$.

Hence, we can only add among ${C,G,H}$ to $CD, GD, HD.$

Adding $C$ to $GD$ and $HD$ we get $GCD$, $HCD$. Taking closure and we will see they are not candidate keys.

Adding H to GD we get GHD which is also not a candidate key.(no more options with $3$ attributes possible)

Now we need to check for candidate keys with $4$ attributes. Since, only remaining options are $CGH$ and we have to add $D$ only possible key of size $4$ is $CGHD$ whose closure also doesn't give us all of the attributes in the relation (All possible options covered)

Hence, no of candidate keys are $4$ : **AD,BD,ED,FD.**

Correct Answer: $B$