Equilibrium: Effect of change in composition and value of Kp, Kc on changing pressure
$K_p$ will not change with pressure, only with temperature. $Q_p$ will instantaneously increase upon applying the pressure for this reaction (why? see below), and will drive the reaction to reactants to re-achieve equilibrium (decreasing $Q_p$) until $Q_p$ = $K_p$ once again. From a LeChatelier approach, you  have more gas molecules in the products, so an increase in pressure will drive the equilibrium composition to the reactants.
As for why $K_p$ won't change with pressure:
$\ln K_p=-\frac{\Delta G^⦵}{RT}$ is at a specified $p^⦵$ (usually 1 bar), so $\left(\frac{\partial {\ln}K_p}{\partial P}\right)_T = 0 $.
And as for why $K_c$ won't change with pressure, see the answer here