It will depend on the state of the product.And if the product is a gas - then the no. of molceules of the gas.
Basically the no. of molcules of gas in each side of the equation.
If more no. of gas molcules on the right hand side - increasing the pressure will shift the equilibrium to the left as less no. of gas molcules.Opposite effect for decrease in pressure
If more no. of gas molcules on the left hand side - increasing the pressure will shift the equilibrium to the right. Opposite effect for decrease in pressure.
If both the sides of the equation have the same no. of gas molcules then position of the equilibrium is not affected by a change in pressure.
Suppose CH4.6H20(s) ----> CH4(g) + 6H2O (l)
<----
Reducing the pressure for this reaction will cause the equilibrium to shift to the left.As 1 gas molceule present on the right hand side of the equation but none on the left hand side.
I hope it helps
Warning
LOL Saud posted before I could