Let's look at nitrosobenzene.

In order to be delocalized (participate in resonance), the lone pair on the nitrogen needs to be in a $p$ orbital. As I have draw it above, the lone pair is in a $sp^2$ orbital, and the $\ce{N=O}$ pi bond is conjugated with the ring:

If the lone pair on the nitrogen atom is delocalized, then yes the nitrogen atom is $sp$ hybridized. As a consequence, the $\ce{N=O}$ pi bond is perpendicular to the pi system. It is no longer conjugated:

Either the lone pair or the $\ce{N=O}$ pi bond is conjugated, not both.

If the lone pair is conjugated, the nitroso group can be electron donating by resonance:

If the $\ce{N=O}$ bond is conjugated, the nitroso group is electron withdrawing by resonance:

Finally, because the nitrogen is more electronegative than carbon, the nitroso group is electron withdrawing by induction:

It is more likely that the $\ce{N=O}$ bond is conjugated because that gives a higher degree of conjugation: 8 atoms versus 7 atoms.

Let's look at nitrosobenzene.

In order to be delocalized (participate in resonance), the lone pair on the nitrogen needs to be in a $\ce{p}$ orbital. As I have draw it above, the lone pair is in a $\ce{sp^2}$ orbital, and the $\ce{N=O}$ pi bond is conjugated with the ring:

If the lone pair on the nitrogen atom is delocalized, then yes the nitrogen atom is $\ce{sp}$ hybridized. As a consequence, the $\ce{N=O}$ pi bond is perpendicular to the pi system. It is no longer conjugated:

Either the lone pair or the $\ce{N=O}$ pi bond is conjugated, not both.

If the lone pair is conjugated, the nitroso group can be electron donating by resonance:

If the $\ce{N=O}$ bond is conjugated, the nitroso group is electron withdrawing by resonance:

Finally, because the nitrogen is more electronegative than carbon, the nitroso group is electron withdrawing by induction:

It is more likely that the $\ce{N=O}$ bond is conjugated because that gives a higher degree of conjugation: 8 atoms versus 7 atoms.