$O_{2}=4.2 \times 10^{-3} M$

and $N O=2.8 \times 10^{-3} M$

For the given reaction,

$N_{2}(g)+O_{2}(g) \rightleftharpoons 2 N O(g)$

equilibrium constant $K_{C}$ can be written as

$K_{C}=\frac{[N O]^{2}}{\left[N_{2}\right]\left[O_{2}\right]}$

$\therefore K_{C}=\frac{\left(2.8 \times 10^{-3} M\right)^{2}}{\left(3.0 \times 10^{-3} M\right)\left(4.2 \times 10^{-3} M\right)}=0.622$

$\because K_{p}=K_{C} \cdot(R T)^{\Delta n}$

$\Delta n=$ Number of moles of gaseous products number of moles of gaseous reactants

$\Delta n=2-2=0$

$\therefore K_{p}=K_{C^{.}}(R T)^{o}$

$K_{p}=K_{C}$ or, $K_{p}=0.622$ $atm$