At a certain temperature and a total pressure of ${10}^5 \mathrm{Pa}$, iodine vapours contain $40\%$ by volume of iodine atoms at equlibrium $\left[{\mathrm{I}}_{2( \mathrm{g})}\rightleftharpoons 2{\mathrm{I}}_{\left(\mathrm{g}\right)}\right]{\mathrm{K}}_{\mathrm{P}}$ for the equilibrium will be

 $\left[{\mathrm{I}}_{2( \mathrm{g})}\rightleftharpoons 2{\mathrm{I}}_{\left(\mathrm{g}\right)}\right]$

The partial pressure of iodine atoms is $\mathrm{P}=\frac{{10}^5\times 40}{100}=0.4\times {10}^6 \mathrm{Pa}$.

The partial pressure of ${\mathrm{I}}_2$ molecules is ${\mathrm{P}}^{\text{'}}=\frac{{10}^5\times 60}{100}=0.6\times {10}^5\mathrm{P}$ a.

The value of the equilibrium constant is ${\mathrm{K}}_{\mathrm{P}}=\frac{{\mathrm{P}}^2}{{\mathrm{P}}^{\text{'}}}=\frac{{\left(0.4\times {10}^5\right)}^2}{0.6\times {10}^5}=2.67\times {10}^4$.

Hence the correct option is (C) $2.67\times {10}^4$