$\begin{array}{llcc}\text { } & \mathrm{N}_2 \mathrm{O}_4(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_2(\mathrm{~g}) \\ & \text { Initial moles } & 1 & 0 \\ & \text { Moles of equil. } & (1-\alpha) & 2 \alpha\end{array}$
( $\alpha=$ degree of dissociation $)$
Total number of moles at equil.
$$
\begin{aligned}
& =(1-\alpha)+2 \alpha \\
& =(1+\alpha) \\
p_{N_2 \mathrm{O}_4} & =\frac{(1-\alpha)}{(1+\alpha)} \times P \\
p_{N O_2} & =\frac{2 \alpha}{(1+\alpha)} \times P \\
K_P & =\frac{\left(p_{N_O}\right)^2}{p_{N_2 O_4}}=\frac{\left(\frac{2 \alpha}{(1+\alpha)} \times P\right)^2}{\left(\frac{1-\alpha}{1+\alpha}\right) \times P}=\frac{4 \alpha^2 P}{1-\alpha^2}
\end{aligned}
$$

$$
\begin{aligned}
& \text { Given, } K_P=2, P=0.5 \mathrm{~atm} \\
& \therefore \quad K_P=\frac{4 \alpha^2 P}{1-\alpha^2} \\
& =\frac{4 \alpha^2 \times 0.5}{1-\alpha^2} \\
& \alpha=0.707 \approx 0.71 \\
& \therefore \quad \text { Percentage dissociation } \\
& =0.71 \times 100=71
\end{aligned}
$$