Why is nitrogen sp3 hybridized even when it bonds to only three other atoms?

The lone pair on nitrogen is stereochemically active. The electron pairs around nitrogen, 3 bonding, and 1 non-bonding, are tetrahedral to a first approx. #/_H-N-H# is compressed from the ideal tetrahedral angle due to the disproportionate influence of the non-bonding lone pair.

When ammonia is quaternized to give ammonium ion, #NH_4^+#, the geometry about nitrogen is tetrahedral with #/_H-N-H=109.5^@#.

Nitrogen has atomic number 7, so electronic configuration will be: #1s^2 2s^2 2p^3#. However, following Hunds rule, the 3 p electrons occupy 3 degenerate p orbitals (x, y and z).

When the s and p orbitals are hybridised to create sp3 hybrid orbitals, we have 4 hybrid orbitals to be filled with 5 electrons (because nitrogen has 5 valence electrons).

Each of the four sp3 hybrid orbitals takes one electron each, and that leaves us with 1 more electron . This then goes into one of the sp3 hybrid orbitals that already contains 1 electron, giving it a total of 2 electrons. These are often shown as "lone pair" on a Lewis diagram, but they are in fact in a non-bonding orbital (the MO theory 'version' of a lone pair).

Therefore 1 s orbital and 3 p orbitals create four degenerate sp3 orbitals, but bonding occurs to only 3 other atoms, and those in the non-bonding orbital do not participate.