Answer
$\ 4.7\times10^9$
Work Step by Step
All the chemical species are in gaseous state except the C(s) which is solid state , Hence we do not include the $p_{C}$ in the expression for $K_p$
Now, for the first reaction,
$$\ K_{p}' = \frac{P_{CO}^{2}}{P_{CO_2}} $$
for the Second Reaction,
$$\ K_{p}'' = \frac{P_{COCl_2}}{(P_{Cl_2}\times P_{CO_2}) } $$
for the third reaction ,
$$\ K_{p} = \frac{P_{COCl_2}^{2}}{(P_{Cl_2}^2\times P_{CO_2})} $$
Now,
$$K_p'\times{K_p''}^2 = \frac{P_{CO}^{2}}{P_{CO_2}}\times{(\frac{P_{COCl_2}}{(P_{Cl_2}\times P_{CO_2}) })}^2 $$
$$ = \frac{P_{COCl_2}^{2}}{(P_{Cl_2}^2\times P_{CO_2})} = K_p $$
For the numerical value of $ K_p $, we substitute the value of $ K_p'$ and $K_p''$ in the above equation
$$ K_p = K_p'\times K_p''^2 $$
$$ K_p = 1.3\times10^{14}\times(6.0\times10^{-3})^2 $$
$$ K_p = 4.68\times10^9 $$
$$ K_p \approx 4.7\times10^9 $$
