Setting up the van't Hoff relation

For the equilibrium $\ce{NH4+ <=> NH3 + H+}$, thermodynynamics gives:

$$\ln K_a = -\dfrac{\Delta H^\circ}{R}\left(\dfrac{1}{T}\right) + \dfrac{\Delta S^\circ}{R}$$

This is a straight line in $\ln K_a$ vs. $1/T$, with slope $= -\Delta H^\circ/R$ and y-intercept $=\Delta S^\circ/R$.

Reading the graph

The plotted line has a negative slope, so $-\Delta H^\circ/R < 0$, meaning $\Delta H^\circ > 0$: the dissociation of $\ce{NH4+}$ is endothermic. Endothermic dissociation means $K_a$ increases as $T$ increases (since as $T$ increases, $1/T$ decreases, and with a negative slope the line moves up to higher $\ln K_a$).

Statement I

Because $K_a$ increases with temperature, a 0.100 M $\ce{NH4NO3}$ solution dissociates more at higher $T$, producing more $\ce{H+}$. This makes the solution more acidic, so pH decreases, not increases, as $T$ rises. Statement I is false.

Statement II

The y-intercept of the graph ($=\Delta S^\circ/R$) for this reaction is negative — extrapolating the line to $1/T \to 0$ gives a negative $\ln K_a$ intercept. This reflects the fact that although one mole of ions/molecules becomes two, the reaction converts a charged, highly hydrated ($\ce{NH4+}$) species and forms $\ce{H+}$ (also strongly hydrated) plus neutral $\ce{NH3}$; the increased ordering of solvent water around the separated charges outweighs the increase in translational entropy from more particles. Thus $\Delta S^\circ$ is actually negative, not positive. Statement II is false.

Conclusion: Neither I nor II is correct — answer (D).