(A) The expression for the equilibrium constant $K_p$ is:
$K_p = \frac{p_{CO} \times p_{H_2}^3}{p_{CH_4} \times p_{H_2O}}$
$(b)$ $(i)$ Increasing the pressure shifts the equilibrium in the backward direction because the number of moles of gaseous products $(4 \ mol)$ is greater than the number of moles of gaseous reactants $(2 \ mol)$. $K_p$ remains unchanged as it depends only on temperature.
$(ii)$ Increasing the temperature shifts the equilibrium in the forward direction because the reaction is endothermic $(\Delta H > 0)$. Consequently,the value of $K_p$ increases.
$(iii)$ Using a catalyst does not affect the equilibrium position or the value of $K_p$. It only increases the rate at which the equilibrium is attained.