Okay, let's see.
This is asking you to consider which of nitrogen's valence electrons count towards the #2n+2# rule (Hückel's Rule) when it comes to aromaticity.
The only ones that are delocalized within the #pi# electron cloud are the ones in the ring, i.e. not in an orbital perpendicular to the #2p_z# (vertical) orbitals of each carbon in the ring.
There are only three electron groups around each nitrogen, indicating a trigonal planar electron geometry.
Since the #sp^2# orbitals are in the same plane as the rest of the molecule, and since the #z# axis is vertical and the #xy#-plane is horizontal, the #xy#-plane is perpendicular to the #z# axis; so, the #sp^2# orbital of nitrogen is perpendicular to the #2p_z# of each carbon.
Hence, the #sp^2# orbital does not contribute its lone pair into the ring, and nitrogen only contributes electrons into the ring that are already bonding.
Each nitrogen therefore contributes approximately three #pi# electrons to the aromatic ring. If we assume the electronegativities of carbon and nitrogen are similar enough, then there would be three #pi# electrons per nitrogen.
