The correct set of four quantum numbers for the valence electron of rubidium atom ($Z=37$) is:

Electronic configuration of Rubidium ($Z=37$):
$$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^1$$
Or $[\text{Kr}]5s^1$.The valence electron is in the $5s$ orbital.For $5s$ orbital:Principal quantum number $n=5$ Azimuthal quantum number $l=0$ (for $s$ orbital)Magnetic quantum number $m=0$ Spin quantum number $s=+1/2$ or $-1/2$ Option D matches this set.