Ammonia contains three bonds where nitrogen is bonded to hydrogen. Nitrogen has a total of five valence electrons. This means that three of those electrons are involved in the bond formation with hydrogen and 2 electrons will be left unbonded. This will lead to a formation of a lone pair. Therefore, we can say that nitrogen has one lone pair of electrons.
Water contains 2 oxygen atoms bonded to hydrogen atoms. Oxygen contains 6 electrons in its valence orbitals which means that out of the 6 electrons, two electrons will be involved in the bonding with hydrogen. This means that 4 electrons will remain. For this reason, we can say that oxygen will have two lone pairs of electrons.
We also know that the presence of either a lone pair or a bond pair causes repulsions in a compound. That is there are repulsions between a lone pair and another lone pair of electrons, between a lone pair and a bond pair and between two bond pairs. This is because of the electron density that is involved in these bonds.
Therefore, we can say that the bond angle in ammonia is more than that of the bond angle of water. Therefore, we can conclude by saying that the assertion that even though central atom of both NH3 and H2O molecules are sp3 C hybridized, yet H−N−H bond angle is greater than that of H−O−H is true. And the reason, that is, nitrogen atom has a lone pair and oxygen atom has two lone pairs is true as well.
Therefore, both the assertion and reason is true and the reason is the correct explanation of the assertion