How does the equilibrium shift when the products are solid and gas when increasing pressure?
Consider the equilibrium expression for this reaction.
$$\ce{CaCO3(s) &lt;=&gt; CaO(s) + CO2(g)}\quad K_c^{'} = \frac{[\ce{CaO}] [\ce{CO2}]}{ [\ce{CaCO3}]}$$
The concentrations of solids and liquids are constant.  They are the molar densities. Since $\ce{[CaO]}$ and $\ce{[CaCO3]}$ don't change, they are moved to the left hand side and &quot;folded into&quot; the equilibrium constant.
\begin{align}
K_c^{'} &amp;\cdot \frac{\ce{[CaCO3]}}{\ce{[CaO]}} = \ce{[CO2]} \\
K_c &amp;= \ce{[CO2]}
\end{align}
Therefore, as long as solid $\ce{CaO}$ and solid $\ce{CaCO3}$ are present along with $\ce{CO2}$ gas there will be an equilibrium.  Only changes to the concentration of $\ce{CO2}$ will cause a shift in the equilibrium.
You asked how will the amounts change if the pressure is increased.  The pressure of $\ce{CO2}$ is increased by either adding more $\ce{CO2}$ or by reducing the volume of the container.  Adding more $\ce{CO2}$ will increase the concentration of $\ce{CO2}$ momentarily, which will shift the equilibrium to the left, using up some $\ce{CaO}$ and making $\ce{CaCO3}$. The pressure of $\ce{CO2}$ can also be increased by reducing the volume of the container.  Again, the concentration of $\ce{CO2}$ is increased, which increases the reaction with $\ce{CaO}$ to make additional $\ce{CaCO3}$.  The $\ce{CO2}$ pressure gets closer to what it was originally, as predicted by Le Chatelier's principle. Since the molar densities of CaO and $\ce{CaCO3}$ are constant, they don't appear in the equilibrium expression. That is why only changes to the pressure (concentration) of $\ce{CO2}$ 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 $\ce{CO2}$, $\ce{CaO}$ or $\ce{CaCO3}$.  The added gas won't affect the partial pressure of $\ce{CO2}$.