Reaction $2Ba{O}_2\left(s\right)\iff 2BaO\left(s\right)+O_2\left(g\right),\Delta H=+ve$. At equilibrium condition, pressure of $O_2$ is depended on:

$Ba{O}_2\left(s\right)\stackrel{r_1}{\underset{r_2}{\iff }}BaO\left(s\right)+O_2\left(g\right),\Delta H=+ve$
According to law of mass action, the rate of forward reaction $=r_1$
$r_1\propto \left[Ba{O}_2\right]$
or $r_1=k_1\left[Ba{O}_2\right]$
$Ba{O}_2$ is solid substence in pure state concentration $=1m$.
then, $r_{=}{k}_1$
Similarly the rate of backward reaction $=r_2$
$r_2\propto \left[BaO\right]\left[O_2\right]$
or $r_2=k_2\left[BaO\right]\left[O_2\right]$
$\because$ Concentration of solid $\left[BaO\right]=1\left[O_2\left(g\right)\right]$
$\therefore r_2=k_2\left[O_2\right]$
At reuilibrium ,
$r_1=r_2$
$K_1=K_2\left[O_2\right]$
or $K_1=K_2.p_{\odot 2}$
where $\left(o{.}_2\right)$ =partial pressure of $O_2$
or $\frac{K_1}{K_2}=p_{o{.}_2}\left(\text{equilibrium constant}\right)$
$\because \frac{K_1}{K_2}=K$ or $K=p_{\odot 2}$