The decreasing order of second ionization potential of $\mathrm{K},\mathrm{Ca},\mathrm{Ba}$ is

Second ionization energy is the energy that is required in the removal of an electron when one electron has already been removed from the atom. Considering their electronic configurations,

$\mathrm{K}=\left[\mathrm{Ar}\right]4{\mathrm{s}}^1$

$\mathrm{Ca}=\left[\mathrm{Ar}\right]4{\mathrm{s}}^2$

$\mathrm{Ba}=\left[\mathrm{Xe}\right]6{\mathrm{s}}^2$

In case of potassium, the second electron will be removed from the $3\mathrm{p}$ subshell which is full-filled and stable. Hence, more energy will be required to remove an electron. 

In case of calcium, the removal of electrons in second I.E occurs from the 4s subshell which is half-filled. Hence, less energy will be required in comparison to potassium. 

And in case of barium, the second electron will be removed from the $6\mathrm{s}$ subshell which is half-filled and easy removal of electrons occurs. So, very less energy is required.

Thus, the decreasing order of the second I.E is $\mathrm{K}>\mathrm{Ca}>\mathrm{Ba}$.

Therefore, option (A) is correct.