According to MO theory the bond orders for O$$_2^{2-}$$, CO and NO$$^+$$ respectively, are

The bond order is calculated by:

$$Bond\ Order\ =\ \frac{N_b-N_a}{2}$$

where,

$$N_b$$ is number of bonding electrons

$$N_a$$ is number of antibonding electrons

For, $$O_2^{2-}$$

Total electrons = 8+8+2 = 18 electrons

MO Configuration:
$$\sigma_{1s}^2 \ \sigma_{1s}^{*2} \ \sigma_{2s}^2 \ \sigma_{2s}^{*2} \ \sigma_{2p_z}^2 \ (\pi_{2p_x}^2 = \pi_{2p_y}^2) \ (\pi_{2p_x}^{*2} = \pi_{2p_y}^{*2})$$
$$N_{b} = 10$$, $$N_{a} = 8$$
$$\text{Bond Order} = \frac{10 - 8}{2} = 1$$

For, $$CO$$
Total Electrons: $6 + 8 = 14\text{ electrons}$
MO Configuration: $$\sigma_{1s}^2 \ \sigma_{1s}^{*2} \ \sigma_{2s}^2 \ \sigma_{2s}^{*2} \ (\pi_{2p_x}^2 = \pi_{2p_y}^2) \ \sigma_{2p_z}^2$$
$$N_b = 10$$, $$N_a = 4$$
$$\text{Bond Order} = \frac{10 - 4}{2} = 3$$

For, $$\text{NO}^+$$
Total Electrons: $7 + 8 - 1 = 14\text{ electrons}$
MO Configuration: Since it is isoelectronic with $\text{CO}$, it shares the same molecular orbital distribution.
$$N_b = 10$$, $$N_a = 4$$
$$\text{Bond Order} = \frac{10 - 4}{2} = 3$$

Hence, Option A is correct.