Pressure only affects a reversible reaction at equilibrium if gases are involved. The key is to count the number of gas molecules on each side of the balanced symbol equation. The number of molecules is the same as the number of moles shown by the balancing numbers.
If the pressure is increased, the equilibrium position shifts towards the side with the smaller number of molecules. This lowers the pressure again. If the pressure is decreased, the equilibrium position shifts towards the side with the larger number of molecules, which raises the pressure again.
Take the reaction N2(g) + 3H2(g) ⇌ 2NH3(g). The left-hand side has 1 + 3 = 4 molecules. The right-hand side has 2 molecules. A higher pressure shifts the position to the right, so the relative amount of ammonia at equilibrium increases. A lower pressure shifts it to the left, so less ammonia is present.
Now take N2O4(g) ⇌ 2NO2(g). There is 1 molecule on the left and 2 molecules on the right. A higher pressure shifts the position to the left. If both sides have the same number of molecules, a change in pressure has no effect on the equilibrium position.