Decomposition of CaCO3 in a sealed vessel
You are right that at $1100~\mathrm{K}$ there will not be any decomposition of $\ce{CaCO3}$ in the presence of $\ce{CO2}$.

But, as the temperature increases further to $1120~\mathrm{K}$, $\ce{CaCO3}$ starts decomposing because at this temperature $Q_\mathrm p &lt; K_\mathrm p$.

As the reaction is taking place in a closed container, by viewing the reaction mixture as an ideal gas mixture, we can calculate the amount of $\ce{CaCO3}$ decomposed at $1120~\mathrm{K}$:

$$\mathrm P(\ce{CO2}) \times V = 1 \times R \times 1100 \tag1$$
$$\mathrm P(\ce{CO2}) = 0.80$$
$$\mathrm P(\ce{CO2}) \times V = (1+n) \times R \times 1120 \tag2$$
$$\mathrm P(\ce{CO2}) = 1.14$$

On solving, $n= 0.399$.