(i) X ( g ) ⇌ Y ( g ) + Z ( g ) K _ p1 = 3 (ii) A ( g ) ⇌ 2B ( g )...

$X ( g ) ⇌ Y ( g ) + Z ( g ) {K}_{{p}_{1}} = 3$

$n$ Initial concentration

$n - \alphan\alphan\alphan$ at equilibrium
Let total pressure be ${p}_{1}$

Using dalton's law of partial pressures and equation for equilibrium constant:
$∴ {K}_{{p}_{1}} = \dfrac{{\left(\dfrac{\alpha}{1 + \alpha} \times {p}_{1}\right)}^{2}}{\dfrac{1 - \alpha}{1 + \alpha}{p}_{1}}$

$⇒ 3 = \dfrac{{\alpha}^{2} \times {p}_{1}}{1 - {\alpha}^{2}}$ ....(X)

$A ( g ) ⇌ 2B ( g ) {k}_{{p}_{2}} = 1$

$n$ Initial mole

$x - \alpha n2\alpha n$ at equilibrium
Let total pressure be ${p}_{2}$
Using dalton's law of partial pressures and equation for equilibrium constant

$∴ {k}_{{p}_{2}} = \dfrac{{\left(\dfrac{2\alpha}{1 + \alpha} \times {p}_{2}\right)}^{2}}{\dfrac{1 - \alpha}{1 + \alpha} \times {p}_{2}}$

$⇒ 1 = \dfrac{4{\alpha}^{2} \times {p}_{2}}{1 - {\alpha}^{2}}$ ....(Y)
Thus, dividing equation (X) by equation(Y)

$⇒ \dfrac{{k}_{{p}_{1}}}{{k}_{{p}_{2}}} = \dfrac{{p}_{1}}{4{p}_{2}}$

$⇒ \dfrac{3}{1} = \dfrac{{p}_{1}}{4{p}_{2}}$

$∴ \dfrac{{p}_{1}}{{p}_{2}} = 12$