Explanation

(d)
$NO(7 + 8 = 15) = \sigma 1s^{2},\sigma^{*}1s^{2},\sigma 2s^{2},\sigma^{*}2s^{2},\sigma 2p_{\mathcal{z}}^{2},\pi 2p_{x}^{2} \approx \pi 2p_{y}^{2},\pi^{*}2p_{x}^{1}$
$NO^{+}(7 + 8 - 1 = 14) = \sigma 1s^{2},\sigma^{*}1s^{2},\sigma 2s^{2},\sigma^{*}2s^{2},\sigma 2p_{\mathcal{z}}^{2},\pi 2p_{x}^{2} \approx \pi 2p_{y}^{2}$
Thus, in the formation of $NO^{+}$ from NO, the electron is removed from a $\pi^{*}$orbital