The equilibrium constants $$K_{P_1}$$ and $$K_{P_2}$$ for the reactions $$X \rightleftharpoons 2Y$$ and $$Z \rightleftharpoons P + Q$$, respectively are in the ratio of 1 : 9. If the degree of dissociation of $$X$$ and $$Z$$ be equal then the ratio of total pressure at these equilibria is

Consider 1 mol of each reactant initially and let the common degree of dissociation be $$\alpha$$.

Case 1: $$X \rightleftharpoons 2Y$$

At equilibrium the number of moles are
  $$X : 1-\alpha,\; Y : 2\alpha,\; n_{\text{total}} = 1+\alpha$$

Total pressure = $$P_1$$, therefore partial pressures are
  $$p_X = \frac{1-\alpha}{1+\alpha}\,P_1,\qquad p_Y = \frac{2\alpha}{1+\alpha}\,P_1$$

For $$X \rightleftharpoons 2Y$$, $$K_{P_1} = \dfrac{p_Y^{\,2}}{p_X}$$

$$K_{P_1} = \dfrac{\left(\dfrac{2\alpha}{1+\alpha}P_1\right)^{2}} {\dfrac{1-\alpha}{1+\alpha}P_1} = \dfrac{4\alpha^{2}P_1}{(1+\alpha)(1-\alpha)} = \dfrac{4\alpha^{2}P_1}{1-\alpha^{2}} \; -(1)$$

Case 2: $$Z \rightleftharpoons P + Q$$

At equilibrium the number of moles are
  $$Z : 1-\alpha,\; P : \alpha,\; Q : \alpha,\; n_{\text{total}} = 1+\alpha$$

Total pressure = $$P_2$$, hence
  $$p_Z = \frac{1-\alpha}{1+\alpha}\,P_2,\qquad p_P = p_Q = \frac{\alpha}{1+\alpha}\,P_2$$

For $$Z \rightleftharpoons P + Q$$, $$K_{P_2} = \dfrac{p_P\,p_Q}{p_Z}$$

$$K_{P_2} = \dfrac{\left(\dfrac{\alpha}{1+\alpha}P_2\right)^{2}} {\dfrac{1-\alpha}{1+\alpha}P_2} = \dfrac{\alpha^{2}P_2}{(1+\alpha)(1-\alpha)} = \dfrac{\alpha^{2}P_2}{1-\alpha^{2}} \; -(2)$$

Divide (1) by (2):

$$\frac{K_{P_1}}{K_{P_2}} = \frac{4\alpha^{2}P_1}{1-\alpha^{2}}\Big/\frac{\alpha^{2}P_2}{1-\alpha^{2}} = \frac{4P_1}{P_2}$$

The problem states $$K_{P_1} : K_{P_2} = 1 : 9$$, i.e. $$\dfrac{K_{P_1}}{K_{P_2}} = \dfrac{1}{9}$$.

Hence
$$\frac{1}{9} = \frac{4P_1}{P_2} \quad\Longrightarrow\quad \frac{P_1}{P_2} = \frac{1}{36}$$

Therefore the ratio of total pressures is $$P_1 : P_2 = 1 : 36$$.

Option A which is: 1 : 36