The correct order of second ionization energy amongst K, Ca and Ba is

The second ionization implies the following facts.

$\begin{array}{l}{ }_{19}\mathrm{K}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\left(3\mathrm{s}{)}^2\right(3\mathrm{p}{)}^6\to \left(3\mathrm{s}{)}^2\right(3\mathrm{p}{)}^5\\ { }_{20}\mathrm{Ca}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\left(3\mathrm{s}{)}^2\right(3\mathrm{p}{)}^6(4\mathrm{s}{)}^1\to (3\mathrm{s}{)}^2(3\mathrm{p}{)}^6\\ { }_{56}\mathrm{Ba}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\left(5\mathrm{s}{)}^2\right(5\mathrm{p}{)}^6(6\mathrm{s}{)}^1\to (5\mathrm{s}{)}^2(5\mathrm{p}{)}^6\end{array}$

of these, the second ionization energy will be largest for K due to the removal of electron from the fully-filled stable ${(3\mathrm{p} )}^6$ configuration.