The given reaction is a heterogeneous equilibrium: $\text{BaO}_2(s) \rightleftharpoons \text{BaO}(s) + \text{O}_2(g)$ For heterogeneous equilibria involving pure solids, their active masses are taken as unity. Thus, the equilibrium constant in terms of partial pressure ($K_p$) is given by: $K_p = P_{\text{O}_2}$ This shows that the equilibrium pressure of $\text{O}_2$ depends only on the value of $K_p$, and is independent of the masses of the solid reactants and products ($\text{BaO}_2$ and $\text{BaO}$). The value of the equilibrium constant ($K_p$) changes only with a change in temperature . Since the reaction is endothermic ($\Delta H = +\text{ve}$), an increase in temperature will increase the value of $K_p$ , which in turn increases the pressure of $\text{O}_2$.