A vessel contain $N_2{O}_4$ and $N{O}_2$  in 3:2 molar ratio at 10 atm under equilibrium now Kp for $N_2{O}_4\rightleftharpoons 2N{O}_2$  is

 $N_2{O}_4\rightleftharpoons 2N{O}_2$
 $K_P=\frac{{\left(P_{N{O}_2}\right)}^2}{P_{N_2{O}_4}}$
$P_T=10atm$ 
$n{N}_2{O}_4=3$ ;  $nN{O}_2=2$
$nt=3+2=5$ 
 $P_{N{O}_2}=X_{N{O}_2}\times P_5$
 $=\frac{2}{5}\times 10=\frac{20}{5}=4$
 $P_{N_2{O}_4}=\frac{3}{5}\times 10=6$
$K_P=\frac{{\left(4\right)}^2}{6}=\frac{16}{6}=2.6atm$