Is is valid to compare reaction quotient to $K_p$ as well as $K_c$?
The reaction of interest is:
$$\ce{XY(s) &lt;=&gt; X(g) + Y(g)} \tag1$$
Thus, $K_p = P_\ce{X}\cdot P_\ce{Y} = 4.1$ since $P_\ce{XY} = 1$ in given condition because it is a solid. If we assume $\ce{X}$ and $\ce{X}$ are real gases, the given conditions are such that $P_\ce{X} = \frac{n_\ce{X} RT}{V} = n_\ce{X}$ and $P_\ce{Y} = \frac{n_\ce{Y} RT}{V} = n_\ce{Y}$, no matter what units are used for $P$.For example:
$$P_\ce{X} = \frac{n_\ce{X} \ \pu{mol} \times  \pu{0.082 L atm K-1 mol-1} \times  \pu{273.15 K}}{\pu{22.4 L}} = n_\ce{X} \ \pu{atm}$$
Suppose, initial amounts of $\ce{X}$ and $\ce{X}$ are $x$ and $y$ and amount of $\alpha$ reacted to produce $\alpha$ amount of  $\ce{XY}$ at the equilibrium (to be realistic, $\alpha \gt 0$). Thus, at the equilibrium: $P_\ce{X} = x- \alpha$ and  $P_\ce{Y} = y- \alpha$. Therefore:
$$K_p = P_\ce{X}\cdot P_\ce{Y} = (x- \alpha)(y- \alpha) = \alpha^2 - (x + y) \alpha + xy = 4.1 \tag2$$

If $x = 5$ and $y = 0.5$: From the equation $(2)$, $$ \alpha^2 - 5.5 \alpha + 2.5 = 4.1 \ \Rightarrow \ \alpha^2 - 5.5 \alpha - 1.6 = 0 \\ \therefore \ \alpha = \frac{5.5 \pm \sqrt{5.5^2 + 4 \times 1.6}}{2} = \frac{5.5 \pm 6.05}{2} $$
Thus, $\alpha = 5.78$ (not realistic) or $\alpha \ne -0.28$ (because $\alpha \gt 0$). Therefore no reaction.
If $x = 2.0$ and $y = 2.0$: Again, from the equation $(2)$, $$ \alpha^2 - 4 \alpha + 4 = 4.1 \ \Rightarrow \ \alpha^2 - 4 \alpha - 0.1 = 0 \\ \therefore \ \alpha = \frac{4 \pm \sqrt{4^2 + 4 \times 0.1}}{2} = \frac{4 \pm 4.05}{2} $$
Thus, $\alpha = 4.025$ (not realistic) or $\alpha \ne -0.025$ (because $\alpha \gt 0$). Therefore no reaction.
If $x = 1.0$ and $y = 1.0$: Similarly, you can prove that there is no reaction.

Hence, answer is &quot;$(d)$. none of the above.&quot;