Explanation

(a)
$\ _{19}K = 1s^{2},2s^{2}2p^{6},3s^{2}3p^{6},4s^{1}$
$\ _{25}Mn = 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}\ 4s^{2}3d^{5}\$
$\ _{28}Ni = 1s^{2}\ 2s^{2}2p^{6}\ 3s^{2}3p^{6}\ 4s^{2}3d^{8}$
$\ _{21}Sc = 1s^{2}\ 2s^{2}2p^{6}\ 3s^{2}3p^{6}\ 4s^{2}3d^{1}$
Therefore, $K$ has least number of electrons in its $M$-shell $(n = 3) = 8$.