The atomic orbital of #"H"# that is both **compatible** and close enough in energy with the #n = 2# atomic orbitals of #"N"# is the #1s#.

The #1s# atomic orbital of #"H"#, with #E = -"13.6 eV"#, is close enough in energy (less than #"12 eV"# away) to the #2p# atomic orbitals of #"N"#, with #E = -"13.1 eV"#, that it can overlap with SOME of them.

In a **nonlinear** molecule, the #2p_y# orbitals of nitrogen line up directly (head-on, colinear) with hydrogen---the #y# direction of the #"N"-"H"# bond is along the bond. The #z# direction is through the tip of the trigonal pyramid.

The MO diagram for #"NH"_3# is:

[Do note though that this diagram has a typo; the #1s# atomic orbital of nitrogen should be a #2s#.]

As you can see, the #2p# atomic orbitals of #"N"# are indeed very close in energy with the hydrogen #1s# atomic orbitals.

The **nonbonding orbital** holding the lone pair is the #3a_1#, which ends up being contributed to mostly by the #2p_z# of #"N"# and slightly by the #1s# orbitals of the three #"H"# atoms as a group.

So, the "orbital overlap" diagram would look something like this: