The decreasing order of second ionisation potential of K, Ca, Ba is

IA 

[M_gas] ns¹ $\underset{-{\mathrm{e}}^{-}}{\overset{+{\mathrm{IE}}_1}{\to }}$[Inert gas(M⁺)] $\underset{-{\mathrm{e}}^{-}}{\overset{+{\mathrm{IE}}_2}{\to }}$
IIA

[M_gas] ns² $\underset{-{\mathrm{e}}^{-}}{\overset{+{\mathrm{IE}}_1}{\to }}$ M⁺ $\underset{-{\mathrm{e}}^{-}}{\overset{+{\mathrm{IE}}_2}{\to }}$ Inert gas[M⁺²]
IE₂ : IA > IIA
Reason :
After removal of 1e^- from IA group elements the formed M⁺ ion gets stable inert gas configuration.Therefore removal of 2^nd  electron from IA group elements is highly difficult and hence 2^nd IE's of IA group elements are greater than that of IIA group elements IE₂
Therefore IE₂ of K > Ca
→ In IIA group from Be to Ba atomic size increases hence IE decreases
Therefore the overall 2^nd I.E's order is
K>Ca>Ba