Setting up the equilibrium expression

For $\ce{CO(g) + H2O(g) <=> CO2(g) + H2(g)}$, the number of moles of gas is the same on both sides ($\Delta n = 0$). If we write $K_p$ in terms of partial pressures,

$$K_p=\dfrac{P_{CO2}\,P_{H2}}{P_{CO}\,P_{H2O}}$$

Since the total pressure is fixed at 1.0 bar, each partial pressure equals (mole fraction) × (1.0 bar), so

$$K_p=\dfrac{x_{CO2}\,x_{H2}}{x_{CO}\,x_{H2O}}$$

This expression contains no explicit total-pressure dependence, so adding $\ce{H2O}$ at constant total $P$ does not simply push the reaction in one obvious direction the way a volume change would for a reaction with $\Delta n \neq 0$.

Why the direction can't be predicted

Adding $\ce{H2O}$ increases $x_{H2O}$ instantaneously, which momentarily makes the reaction quotient $Q_p < K_p$, so the system shifts forward, consuming some $\ce{CO}$ and $\ce{H2O}$ and producing more $\ce{CO2}$ and $\ce{H2}$. However, because the total pressure is re-fixed at 1.0 bar (the container must re-adjust volume), the new equilibrium mole fractions—and therefore the new partial pressure of $\ce{H2}$—depend on the specific initial amounts of all four species and the numerical value of $K$, none of which are given.

Without knowing the initial composition (moles of $\ce{CO}$, $\ce{H2O}$, $\ce{CO2}$, $\ce{H2}$ present before the 5.0 g of water was added) or the value of $K_p$ at 120 °C, we cannot calculate whether $x_{H2}$ (and hence $P_{H2}$) ends up larger, smaller, or unchanged after the shift. The qualitative Le Chatelier shift tells us the reaction moves forward, but the magnitude relative to how much the total pressure redistributes among all species determines whether $P_{H2}$ actually rises or falls — this cannot be pinned down from the information given.

Conclusion: The correct answer is (D) — the effect cannot be determined without knowing the initial composition and/or $K$.