For the reaction $\mathrm{SnO}_2(\mathrm{s}) + 2\mathrm{H}_2(\mathrm{g}) \rightleftharpoons 2\mathrm{H}_2\mathrm{O}(\mathrm{g}) + \mathrm{Sn}(\mathrm{l})$ calculate $\mathrm{K_p}$ at $900\mathrm{K}$, where the equilibrium steam-hydrogen mixture was $45\%$ $\mathrm{H}_{2}$ by volume

For this heterogeneous equilibrium, the pure solid and pure liquid have activity $1$, so only gaseous species appear in $K_p$. $$K_p=\frac{\left(P_{\mathrm{H_2O}}\right)^2}{\left(P_{\mathrm{H_2}}\right)^2}$$ At equilibrium, the gas mixture is $45\%$ $\mathrm{H_2}$ by volume, so the mole fraction of $\mathrm{H_2}$ is $0.45$ and that of $\mathrm{H_2O}$ is $0.55$. $$\frac{P_{\mathrm{H_2O}}}{P_{\mathrm{H_2}}}=\frac{0.55}{0.45}$$ Therefore, $$K_p=\left(\frac{0.55}{0.45}\right)^2$$ $$K_p\approx 1.49$$ This matches option $\mathrm{(A)}$.