1. Paramagnetism criterion

A species is paramagnetic if it has one or more unpaired electrons. It is diamagnetic if all electrons are paired.

So we check the electronic configuration / molecular orbital (MO) configuration of each species.

2. Option A: CO

CO has total electrons: $6 + 8 = 14$ So CO is isoelectronic with $N_2$.

Its MO filling leads to all electrons paired: $\sigma(1s)^2\,\sigma^*(1s)^2\,\sigma(2s)^2\,\sigma^*(2s)^2\,(\pi 2p_x)^2(\pi 2p_y)^2(\sigma 2p_z)^2$

All electrons are paired, so CO is diamagnetic.

3. Option B: $O_2$

$O_2$ has 16 electrons. Its MO configuration contains two unpaired electrons in antibonding $\pi^*$ orbitals: $\pi^*(2p_x)^1\,\pi^*(2p_y)^1$

Hence $O_2$ is paramagnetic.

4. Option C: $B_2$

$B_2$ has total electrons: $5 + 5 = 10$

For lighter diatomic molecules, MO order is: $\sigma(2s),\ \sigma^*(2s),\ \pi(2p_x)=\pi(2p_y),\ \sigma(2p_z)$

After filling, $\pi(2p_x)^1\,\pi(2p_y)^1$ so there are two unpaired electrons.

Hence $B_2$ is paramagnetic.

5. Option D: NO

NO has total electrons: $7 + 8 = 15$ An odd number of electrons means at least one electron remains unpaired. So NO is paramagnetic.

6. Final conclusion

Among the given species, the one that is not paramagnetic is: $\boxed{\text{CO}}$

So the correct option is A.

7. Comparison with stored answer

Stored correct answer: A

My derived answer: A

They match.