For the reaction $X \rightleftharpoons Y + Z$: Initial moles: 1, 0, 0 Moles at equilibrium: $1-\alpha$, $\alpha$, $\alpha$ Total moles at equilibrium = $1 - \alpha + \alpha + \alpha = 1 + \alpha$ Partial pressures: $p_X = \frac{1-\alpha}{1+\alpha} P_1$, $p_Y = \frac{\alpha}{1+\alpha} P_1$, $p_Z = \frac{\alpha}{1+\alpha} P_1$ $K_{p_1} = \frac{p_Y p_Z}{p_X} = \frac{(\frac{\alpha}{1+\alpha} P_1)(\frac{\alpha}{1+\alpha} P_1)}{\frac{1-\alpha}{1+\alpha} P_1} = \frac{\alpha^2}{1-\alpha^2} P_1$

For the reaction $A \rightleftharpoons 2B$: Initial moles: 1, 0 Moles at equilibrium: $1-\alpha$, $2\alpha$ Total moles at equilibrium = $1 - \alpha + 2\alpha = 1 + \alpha$ Partial pressures: $p_A = \frac{1-\alpha}{1+\alpha} P_2$, $p_B = \frac{2\alpha}{1+\alpha} P_2$ $K_{p_2} = \frac{p_B^2}{p_A} = \frac{(\frac{2\alpha}{1+\alpha} P_2)^2}{\frac{1-\alpha}{1+\alpha} P_2} = \frac{4\alpha^2}{1-\alpha^2} P_2$

Given $\frac{K_{p_1}}{K_{p_2}} = \frac{9}{1}$ $\frac{\frac{\alpha^2}{1-\alpha^2} P_1}{\frac{4\alpha^2}{1-\alpha^2} P_2} = \frac{P_1}{4P_2} = \frac{9}{1}$ $\frac{P_1}{P_2} = \frac{36}{1}$ Therefore, the ratio of total pressures at equilibrium is $36 : 1$.