Why sp3 hybridisation in NH3 but bonding with pure p orbitals in case of other hydrides of group 15
The answer to this question is actually much more complicated than most textbooks make it seem. There's a great explanation in Abright, Burdett and Whango "Orbital Interactions in Chemistry" Ch. 9, which I'll try to summarize here. 

As explained in this answer: Why is the inversion barrier larger in PH3 than it is in NH3?, the lone pair orbital in a planar XH3 molecule would be pure $p$. Thus the bonding orbitals average to $sp^2$. As the bond angle decreases from 180, the loss of symmetry makes possible mixing between that lone pair orbital (originally a pure central atom $p$ orbital) and an anti-bonding orbital involving the central atom $s$ orbital. This "s-p mixing" stabilizes the lone pair orbital at the expense of the (empty) antibonding orbital. This mixing necessarily also involves the bonding orbital that involves the central atom $s$ atomic orbital.

In nitrogen, the low energy of the atomic $s$ orbital and the strong overlap interaction of that $s$ orbital with the hydrogen $s$ orbitals means that the bonding orbital made from the atomic $s$ orbital is quite low energy and the antibonding orbital is quite high energy. As a result, mixing with the $p$ lone pair orbital is less favorable, so less mixing occurs and the angle does not deviate as far from 180. 

With the larger central atoms, the interaction of the $s$ orbital with the hydrogen $s$ orbitals is less, so the antibonding orbital is not as high energy. Likewise, the $s$ bonding orbital is not as low energy. So mixing with the lone pair $p$ orbital is more favorable, and the bond angle deviates farther from 180. The figure below (from Albright, figure 9.9) plots the energy of the occupied orbitals. The lowest line represents the $s$ bonding orbital and the top line the lone pair. 



The key point here is that more $s$ character is mixed into the lone pair orbital as you move down from NH3.