How to draw overlapping of pure or hybridized orbitals for Br2 and NO+? Explain the need for the orbital of an atom to hybridized based on the Lewis structures

There is no in either of these diatomic .
Note that in linear diatomic molecules the ##p_z## orbital always points along the internuclear axis so it has to contribute to one of the ##sigma## bonds.
I’ve drawn the overlaps below in the MO diagrams.
BROMINE (HOMONUCLEAR DIATOMIC)
For ##Br_2## it is the simpler of the two examples. It is a homonuclear diatomic so all of its orbitals have a compatible partner: ##4p_x## with ##4p_x## ##4p_y## with ##4p_y## etc.
In this case since there are only two atoms they use their highest energy compatible orbitals to ##sigma## bond on the internuclear axis. The energy of the ##4s## atomic orbital is ##-24.37 eV## and the ##4p## atomic orbitals are ##-12.49 eV## in energy (Inorganic Chemistry Miessler et al. Table 5.2).
Each bromine would donate one ##mathbf(4p_z)## electron to form a ##sigma##-bonding orbital.
As a result there is no orbital hybridization here.
Here is the MO diagram below (I had to draw it myself since I couldn’t find one online; the ##pi_(4px)## and ##pi_(4py)## orbitals—the ##1b_(3u)## and ##1b_(2u)##—are very close in energy to the ##sigma_(4pz)## the ##1b_(1u)##):
Granted that is not ##Br_2##’s highest-occupied molecular orbital (that would be the ##pi_(4px)^*## and ##pi_(4py)^*##—the ##1b_(3u)## and ##1b_(2u)##) but since both the bonding and antibonding ##pi## molecular orbitals are occupied it is the ##sigma_(4p_z)## (##1b_(1u)##) that participates in the ##sigma## bond.
You should notice that the ##1b_(1u)## orbital is the ##sigma_(4pz)## bonding orbital but the ##2b_(1u)##—the ##sigma_(4pz)^*## antibonding orbital—has no electrons so it doesn’t contribute to the ##sigma## bond. If it did ##Br_2## would not exist.
Therefore the ##sigma_(4pz)## indeed is the molecular orbital that represents the single bond on ##Br_2##.
NO##^(mathbf(+))## BONDING (HETERONUCLEAR DIATOMIC)
##NO^(+)## on the other hand is a heteronuclear diatomic. Since it is also diatomic it also does not need to hybridize.
All of nitrogen’s orbitals are compatible with oxygen’s orbitals in energy (and in symmetry but that is less crucial to our understanding for General Chemistry level education).
The MO diagram for neutral ##NO## is as follows (Inorganic Chemistry Miessler et al. Ch. 5 Answer Key):
(I superimposed some orbital depictions on the original diagram and added symmetries and energies.)
If we consider ##NO^(+)## we remove the electron from the highest-occupied molecular orbital so we take out the one from the ##pi_(2px)^*## antibonding orbital (##2b_1##) to form ##NO^(+)##.
At this point its bonds have increased in strength. The bond order changed from:
##(8 – 3)/2 = 2.5##
to:
##(8 – 2)/2 = 3##
So we know it has a triple bond. That means it needs three orbitals contributed from each atom.
There are two electrons in the ##sigma_(2pz)## molecular orbital (##3a_1##) and there are two electrons each in the ##pi_(2px)## (##1b_1##) and ##pi_(2py)## (##1b_2##) molecular orbitals.
##NO^(+)## therefore uses two ##2p_x## atomic orbitals two ##2p_y## atomic orbitals and two ##2p_z## atomic orbitals to bond.
As a result there is no orbital hybridization here.
Each ##sigma##-bonding pair contributes to a ##sigma## bond and each ##pi##-bonding pair contributes to a ##pi## bond. That accounts for the triple bond: one ##sigma## and two ##pi## bonds.