$${\begin{array}{|c|c|c|l|}\hline
\bf{Q_{prev}}& \textbf{D}& \textbf{Q(JK)}&\bf{Explanation} \\\hline
\text{-}&1&0&\text{Now, the D output is 1, meaning J and K = 1; for next cycle} \\\hline 0&0&1& \text{J = K = 1(D output from prev state), so output toggles from 0 to 1} \\ \hline 1&1&1&\text{J = K = 0, so output remains 1} \\ \hline 1&1&0& \text{J = K = 0, so output remains 1} \\ \hline 0&0&1& \text{J = K = 1, so output toggles from 0 to 1}\\ \hline 1&1&1& \text{J = K = 0, so output remains 1} \\ \hline
\end{array}}$$
$\text{D}$ flipflop output will be same as its input and $\text{JK}$ flipflop output toggles when $1$ is given to both $\text{J}$ and $\text{K}$ inputs.
i.e., $Q = D_{\text{prev}}({Q_{\text{prev}}}') + ({D_{\text{prev}}}')Q_{\text{prev}}$
Correct Answer: $A$
