(N/A) The expression for $K_p$ is: $K_p = \frac{(p_{CO})(p_{H_2})^3}{(p_{CH_4})(p_{H_2O})}$
$(b)$ $(i)$ Increasing the pressure: According to Le Chatelier's principle,the equilibrium will shift in the direction that produces fewer moles of gas (the left side),so the amount of $H_2$ will decrease. $K_p$ remains constant.
$(ii)$ Increasing the temperature: Since the reaction is endothermic,increasing the temperature will shift the equilibrium in the forward direction,increasing the yield of $H_2$ and increasing the value of $K_p$.
$(iii)$ Using a catalyst: $A$ catalyst does not affect the value of $K_p$ or the equilibrium composition; it only helps the system attain equilibrium more quickly.