In a closed container, at a fixed given temperature, let's consider the equilibrium:
$N_2O_4(g)$
$2NO_2(g)$
Suppose that the initial partial pressures at equilibrium are the following:
$p_{N_2O_4}=1\;atm$
and
$p_{NO_2}=1\;atm$
Then the total pressure will be equal to $2\;atm$ and the equilibrium constant
$K_p$ $=$ $\frac{p_{NO_2}^2}{p_{N_2O_4}}$ $=$ $\frac{1^2}{1}$ $=$ $1\;atm$
Now compressing the mixture, we double the total pressure to $4\;atm$ :
A new equilibrium will be established, but the constant $K_p$ remains unchanged (same temperature!). Let's calculate the new partial pressures of the two gases: $p_{NO_2}$ $=$ $x\;atm$ and $p_{N_2O_4}$ $=$ $4-x\;atm$ $K_p$ $=$ $\frac{p_{NO_2}^2}{p_{N_2O_4}}$ $1$ $=$ $\frac{x^2}{4-x}$ $x=1.56\;atm$ . (The negative solution of this equation is to be discarded). Now the two partial pressures are no more equal: $p_{NO_2}$ $=$ $1.56\;atm $ and $p_{N_2O_4}$ $=$ $4$ $-$ $1.56$ $=$ $2.44\;atm$
If the total pressure is increased, the equilibrium is displaced to the left, i.e. to the side where the volume is contracted: (2 mole $NO_2$ occupy a bigger volume than 1 mole $N_2O_4$ !)