For the reaction ${\mathrm{A}}_{\left(\mathrm{g}\right)}\rightleftharpoons {\mathrm{B}}_{\left(\mathrm{g}\right)}+{\mathrm{C}}_{\left(\mathrm{g}\right)}, \mathrm{A}$ is $33\%$ dissociated at a total pressure $\mathrm{P}.$ The correct relation between $\mathrm{P}$ and ${\mathrm{K}}_{\mathrm{p}}$ is

Given,

Total pressure$=\mathrm{P}$

Degree of dissociation $=33\% = 0.33$ 

The given equilibrium reaction is,

$\begin{array}{cccccc}& \mathrm{A}\left(\mathrm{g}\right)& \rightleftharpoons & \mathrm{B}\left(\mathrm{g}\right)& +& \mathrm{C}\left(\mathrm{g}\right)\\ \mathrm{at} \mathrm{t}=0& 1& & 0& & 0\\ \mathrm{at} \mathrm{t}={\mathrm{T}}_{\mathrm{eq}}& 1-\mathrm{\alpha }& & \mathrm{\alpha }& & \mathrm{\alpha }\end{array}$

Now, Total number of moles $= 1-\mathrm{\alpha }+\mathrm{\alpha }+\mathrm{\alpha } =1+\mathrm{\alpha }$ 

Partial pressure of $\mathrm{A} =\left(\frac{1-\mathrm{\alpha }}{1+\mathrm{\alpha }}\right)\times \mathrm{P}$

Partial pressure of $\mathrm{B} =\left(\frac{\mathrm{\alpha }}{1+\mathrm{\alpha }}\right)\times \mathrm{P}$

Partial pressure of $\mathrm{C} =\left(\frac{\mathrm{\alpha }}{1+\mathrm{\alpha }}\right)\times \mathrm{P}$

The relation between the ${\mathrm{K}}_{\mathrm{P}}$ and the total pressure is,

${\mathrm{K}}_{\mathrm{P} }=\frac{{\mathrm{P}}_{\mathrm{B}}\times {\mathrm{P}}_{\mathrm{C}}}{{\mathrm{P}}_{\mathrm{A}}}$

Putting all the values of partial pressure, we get

${\mathrm{K}}_{\mathrm{P}} =\frac{{\mathrm{\alpha }}^2\mathrm{P}}{\left(1-{\mathrm{\alpha }}^2\right)}$

${\mathrm{K}}_{\mathrm{P}}=\frac{{\left(0.33\right)}^2\mathrm{P}}{\left[1-{\left(0.33\right)}^2\right]}$

${\mathrm{K}}_{\mathrm{P} }=0.122\mathrm{P}$

$\mathrm{P} =8.19{\mathrm{K}}_{\mathrm{P}}$

$\mathrm{P}\approx 8{\mathrm{K}}_{\mathrm{P}}$

Thus, the relation between $\mathrm{P}$ and ${\mathrm{K}}_{\mathrm{P}}$ is $\mathrm{P} =8{\mathrm{K}}_{\mathrm{P}}$