The bond angle in PH3 is:

N is smaller than P and thus more electronegative. 

Because of this high electronegativity of N, the bond pair of electrons in N-H bond of NH₃ are closer to N than they are in case of P-H bond of PH₃

Since, electron pairs are closer to N, they experience repulsion amongst themselves which moves the N-H bonds more apart from each other as they would be in PH₃.

Therefore, the bond angle in PH₃ is much lesser than NH₃.

Also, to determine the bond angles of hydrides of period 3, we make use of the Drago rule, which explains the exceptionally low angle of hydrides like PH₃.

The Drago’s rule is based on the following conditions:

• The central atom must be of 3rd period or higher.
• There must be atleast one lone pair of electrons on the central atom
• The electronegativity of the surrounding atom must be less than or equal to 2.5

It says that if all of the above conditions are satisfied by a compound, then the energy difference between the atomic orbitals that were expected to participate in hybridization is so high that no mixing of orbitals takes place and hence, no hybridisation.

In such a case, the bonding will take place through pure atomic p-orbitals and thus, bond angle will be 90^O

Since, PH₃ follows all of these conditions, it is known as a Drago compound and hence, its bond angle is , which is exceptionally low when compared with the angle in NH₃.

The correct answer is A: much lesser than NH₃