The combination is equivalent to two capacitors in parallel each with plate area A/2 and separation d. See fig.

Thus,  ${\mathrm{C}}_1=\frac{{\mathrm{K}}_1{\mathrm{\epsilon }}_{\mathrm{e}}(\mathrm{A}/2)}{\mathrm{d}}=\frac{{\mathrm{K}}_1{\mathrm{\epsilon }}_0\mathrm{A}}{2\mathrm{d}}$
           ${\mathrm{C}}_2=\frac{{\mathrm{K}}_2{\mathrm{\epsilon }}_0(\mathrm{A}/2)}{\mathrm{d}}=\frac{{\mathrm{K}}_2{\mathrm{\epsilon }}_0\mathrm{A}}{2\mathrm{d}}$
The equivalent capacitance C is given by
$\begin{array}{r}\mathrm{C}={\mathrm{C}}_1+{\mathrm{C}}_2=\frac{{\mathrm{K}}_1{\mathrm{\epsilon }}_0\mathrm{A}}{2\mathrm{d}}+\frac{{\mathrm{K}}_2{\mathrm{\epsilon }}_0\mathrm{A}}{2\mathrm{d}}\\ =\frac{{\mathrm{\epsilon }}_0\mathrm{A}}{\mathrm{d}}\left(\frac{{\mathrm{K}}_1+{\mathrm{K}}_2}{2}\right)\end{array}$