Since we only have to deal with $2$ variables ($a$ and $b$), solving this with truth table is feasible.
Truth table for $X$ will be:
$$\begin{array}{|c|c|c|c|c|} \hline \textbf{a} & \textbf{b}& \textbf{b} \Rightarrow \textbf{a} &\textbf{a} \Rightarrow \textbf{b} & (\textbf{b} \Rightarrow \textbf{a}) \Rightarrow (\textbf{a} \Rightarrow \textbf{b}) \\\hline \text{0} & \text{0}& \text{1} & \text{1} & \text{1}\\\hline \text{0} & \text{1}& \text{0} & \text{1} & \text{1}\\\hline \text{1} & \text{0}& \text{1} & \text{0} & \text{0}\\\hline \text{1} & \text{1}& \text{1} & \text{1} & \text{1} \\\hline \end{array}$$
So, clearly, $X$ is satisfiable but not a tautology.
Truth table for $Y$ will be:
$$\begin{array}{|c|c|c|c|} \hline \textbf{a} & \textbf{b}& \textbf{a} \Rightarrow \textbf{b} & (\textbf{a} \Rightarrow \textbf{b}) \wedge \textbf{b} \\\hline \text{0} & \text{0}& \text{1} & \text{0}\\\hline \text{0} & \text{1}& \text{1} & \text{1}\\\hline \text{1} & \text{0}& \text{0} & \text{0}\\\hline \text{1} & \text{1}& \text{1} &\text{1}\\\hline \end{array}$$
So, $Y$ is also satisfiable but not a tautology.
This gives the OPTION (D) as the correct answer.