Both statements are true, based on how the electronegativity and size of the atoms bonded to nitrogen or phosphorus affect the lone-pair/bonding-pair repulsions.

Statement I: $\ce{NH3}$ vs $\ce{NF3}$
- Both have a central N with one lone pair and three bonding pairs, so both are trigonal pyramidal with an idealized angle near $109.5^{\circ}$.
- Fluorine is much more electronegative than hydrogen, so in $\ce{NF3}$ the N–F bonding electrons are pulled away from N toward F. This leaves N with more electron density concentrated in the lone pair, which then exerts greater repulsion on the bonding pairs, compressing the F–N–F angle.
- Result: bond angle $\ce{NH3}$ ($\approx 107^{\circ}$) > bond angle $\ce{NF3}$ ($\approx 102^{\circ}$). Statement I is correct.

Statement II: $\ce{NH3}$ vs $\ce{PH3}$
- Moving down the group from N to P, the central atom becomes larger and uses orbitals with more $s$/$p$ character difference (less effective hybridization), so the bonding pairs sit farther from the nucleus and closer to pure $p$ orbital character (which prefer $90^{\circ}$).
- This causes the H–P–H angle to shrink more than the H–N–H angle.
- Result: bond angle $\ce{NH3}$ ($\approx 107^{\circ}$) > bond angle $\ce{PH3}$ ($\approx 93^{\circ}$). Statement II is correct.

Since both I and II are true, the answer is (C).