The bond angle H-X-H is the greatest in the compound:

First, we recall VSEPR theory, which predicts molecular shapes by counting $$\text{bond pairs}$$ and $$\text{lone pairs}$$ on the central atom. The rule states that regions of electron density repel each other and arrange to minimize repulsion, deciding both shape and bond angles.

We list each molecule, the central atom, its valence electrons, the number of bonding pairs with hydrogen, and the number of lone pairs.

For $$PH_3$$ we have: $$P$$ is in group 15, so $$5$$ valence electrons. Bonding with three $$H$$ atoms uses $$3$$ pairs, leaving $$5-3=2$$ electrons, i.e. $$1$$ lone pair. Thus there are $$4$$ regions of electron density (3 bond pairs + 1 lone pair), giving a trigonal-pyramidal arrangement. By experience and experimental data, the bond angle is close to $$93.5^{\circ}$$ because the larger, more diffuse $$3p$$ orbitals of phosphorus do not overlap strongly with $$1s$$ of hydrogen, so bond-bond repulsion is weak.

For $$CH_4$$ we have: $$C$$ is in group 14 with $$4$$ valence electrons. It forms $$4$$ sigma bonds with $$H$$, using all its electrons and leaving $$0$$ lone pairs. Thus there are exactly $$4$$ regions of electron density, all bond pairs. According to VSEPR, $$4$$ bond pairs adopt a regular tetrahedral geometry. The standard tetrahedral bond angle is $$\displaystyle 109^{\circ}28'$$ (commonly written as $$109.5^{\circ}$$).

For $$NH_3$$ we have: $$N$$ is in group 15 with $$5$$ valence electrons. After making $$3$$ N-H bonds, $$5-3=2$$ electrons remain as $$1$$ lone pair. Again we get $$4$$ regions of electron density (3 bond pairs + 1 lone pair), giving a trigonal-pyramidal shape. However, nitrogen’s $$2p$$ orbitals are small and overlap well with hydrogen’s $$1s$$, producing stronger bond-bond repulsion than in $$PH_3$$. Still, the lone pair exerts extra repulsion and compresses the H-N-H angle to about $$107^{\circ}$$, which is less than the tetrahedral value.

For $$H_2O$$ we have: $$O$$ belongs to group 16 with $$6$$ valence electrons. Two electrons form two O-H bonds, so $$6-2=4$$ electrons remain as $$2$$ lone pairs. Now the total is $$4$$ regions of electron density (2 bond pairs + 2 lone pairs), leading to a bent (angular) geometry. Two lone pairs repel even more strongly, squeezing the H-O-H angle down to about $$104.5^{\circ}$$.

Now we compare the numerical bond angles obtained:

$$PH_3 : \approx 93.5^{\circ}$$

$$NH_3 : \approx 107^{\circ}$$

$$H_2O : \approx 104.5^{\circ}$$

$$CH_4 : 109.5^{\circ}$$

The greatest of these is clearly the tetrahedral angle $$109.5^{\circ}$$ in $$CH_4$$.

Hence, the correct answer is Option B.