Consider the equilibrium expression for this reaction.
Kc' = [CaO] [CO2] / [CaCO3]
The concentrations of solids and liquids are constant. They are the molar densities. Since [CaO] and [CaCO3] don't change, they are moved to the left hand side and "folded into" the equilibrium constant.
Kc' x [CaCO3] / [CaO] = [CO2]
Kc = [CO2]
Therefore, as long as solid CaO and solid CaCO3 are present along with CO2 gas there will be an equilibrium. Only changes to the concentration of CO2 will cause a shift in the equilibrium.
You asked how will the amounts change if the pressure is increased. The pressure of CO2 is increased by either adding more CO2 or by reducing the volume of the container. Adding more CO2 will increase the concentration of CO2 momentarily, which will shift the equilibrium to the left, using up some CaO and making CaCO3. The pressure of CO2 can also be increased by reducing the volume of the container. Again, the concentration of CO2 is increased, which increases the reaction with CaO to make additional CaCO3. The CO2 pressure gets closer to what it was originally, as predicted by Le Chatelier's principle. Since the molar densities of CaO and CaCO3 are constant, they don't appear in the equilibrium expression. This is why only changes to the pressure (concentration) of CO2 affect the position of the equilibrium.
If the pressure in the container is increased by adding an inert or non-reacting gas, nothing happens to the amounts of CO2, CaO or CaCO3. The added gas won't affect the partial pressure of CO2.