Explanation

(b)
The paramagnetic species has unpaired electron in it.
(a)$H_{2} = 1 + 1 = \sigma 1s^{2}$
(b)$N_{2} = 7 + 7 = 14 = \ \sigma 1s^{2},\ _{\sigma}^{*}1s^{2}\ ,\sigma 2s^{2},\ _{\sigma}^{*}2s^{2}\ ,\ \pi 2p_{x}^{2} = \pi 2p_{y}^{2},\sigma 2p_{z}^{2}$
(c)$CO = 6 + 8 = 14 = \ \sigma 1s^{2},\ _{\sigma}^{*}1s^{2}\ ,\sigma 2s^{2},\ _{\sigma}^{*}2s^{2}\ ,\ \pi 2p_{x}^{2} = \pi 2p_{y}^{2},\sigma 2p_{z}^{2}$
(d)$O_{2} = 8 + 8 = 16 = \ \sigma 1s^{2},\ _{\sigma}^{*}1s^{2}\ ,\sigma 2s^{2},\ _{\sigma}^{*}2s^{2}\ ,\ \sigma 2p_{z}^{2},\left( \pi 2p_{x}^{2} = \pi 2p_{y}^{2} \right),(\ _{\pi}^{*}2p_{x}^{1} = \ _{\pi}^{*}2p_{y}^{1})$
$\because$ $O_{2}$ molecule has unpaired electrons.
$\therefore$ $O_{2}$ molecule is paramagnetic.