Step 1: Write the given rule as a formula.
"A Creative person succeeds if and only if disciplined" (with succeeding being impossible for creative-but-undisciplined people) translates to the axiom $A: C \Rightarrow (S \Leftrightarrow D)$. Any world we consider must make $A$ true, since it is given as a fact about how these three properties relate.
Step 2: Go through all 8 combinations of $C, S, D$ and mark which satisfy $A$.
$C{=}0$: $A$ is automatically true no matter what $S, D$ are, since an implication with a false antecedent is always true. That covers all 4 rows with $C{=}0$.
$C{=}1$: $A$ needs $S \Leftrightarrow D$, so only $(S{=}0,D{=}0)$ and $(S{=}1,D{=}1)$ satisfy it; the rows $(S{=}0,D{=}1)$ and $(S{=}1,D{=}0)$ with $C{=}1$ violate $A$ and are excluded.
So the 6 surviving rows are: $(0,0,0), (0,0,1), (0,1,0), (0,1,1), (1,0,0), (1,1,1)$, written as $(C,S,D)$.
Step 3: Check statement (ii) is just $A$ itself.
Statement (ii) is $C \Rightarrow (S \Leftrightarrow D)$, which is identical to $A$. Since we only ever consider worlds where $A$ holds, statement (ii) is true in every one of the 6 surviving rows, so it is always TRUE.
Step 4: Check statement (i) on the row $(0,0,1)$.
Here $C \wedge S = 0 \wedge 0 = 0$, while $D = 1$. So $C \wedge S \Leftrightarrow D$ becomes $0 \Leftrightarrow 1$, which is false. Statement (i) fails on this valid row, so it is not always true.
Step 5: Check statement (iii) on the row $(0,0,0)$.
Here $D \Rightarrow S = 0 \Rightarrow 0 = 1$ and $\neg S = 1$, so $(D \Rightarrow S) \vee \neg S = 1$. But $C = 0$, so $C \Leftrightarrow 1$ is false. Statement (iii) fails on this valid row too, so it is not always true.
Final Answer:
Only statement (ii) is guaranteed true across every world consistent with the rule, so the answer is option (B).