Equilibrium for the synthesis of methanol Liquid methanol is obtained with carbon monoxide gas and hydrogen gas in a reactor at $\pu{300 °C}$ and $\pu{250 atm}.$ $K_p = \pu{9.28E-3}.$ Find $K_c.$

With only these data, do I have to use the following equilibrium for the calculations
$$\ce{CO(g) + 2 H2(g) &lt;=&gt; CH3OH(l)},$$
or do I have to suppose first methanol is obtained as a gas and then liquified:
$$\ce{CO(g) + 2 H2(g) &lt;=&gt; CH3OH(g)}?$$
Note that under the given conditions, methanol cannot exist in the liquid state, however since you are given in the problem that methanol is obtained in liquid state, you will only consider the equilibrium (for problem solving purpose)
$$\ce{CO(g) + 2 H2(g) &lt;=&gt; CH3OH(l)}$$
And since $K_p$ is related to $K_c$ by the relation
$$K_p = K_c(RT)^{\Delta n_\mathrm{g}},$$ where $\Delta n_\mathrm{g}$ represents the change in the number of moles of gaseous species, so here $\Delta n_\mathrm{g} = -3,$ Therefore $K_\mathrm{c}=K_\mathrm{p}(RT)^3=9.28\times 10^{-3}\times (300 \times 0.0821)^3 \approx \pu{138.65 atm^3 L^3 mol^-3}$.