At a certain temperature and total pressure of 105

Partial pressure of I atoms,

$p_{1}=\frac{40}{100} \times p_{\text {total }}$

$=\frac{40}{100} \times 10^{5}$

$=4 \times 10^{4} \mathrm{~Pa}$

Partial pressure of I2 molecules,

$p_{\mathrm{l}_{2}}=\frac{60}{100} \times p_{\text {total }}$

$=\frac{60}{100} \times 10^{5}$

$=6 \times 10^{4} \mathrm{~Pa}$

Now, for the given reaction,

$K_{p}=\frac{(p \mathrm{I})^{2}}{p_{1_{z}}}$

$=\frac{\left(4 \times 10^{4}\right)^{2} \mathrm{~Pa}^{2}}{6 \times 10^{4} \mathrm{~Pa}}$

$=2.67 \times 10^{4} \mathrm{~Pa}$