A parallel plate capacitor with square plates is filled with four dielectrics of dielectric constants  $K_1,K_2,K_3,K_4$ arranged as shown in the figure. The effective dielectric constant  $K$ will be:

This capacitor system can be converted into two parts as sown in the figure

Where $C_1,C_2,C_3$ and $C_4$ are capacitance of the capacitor having dielectric constants $K_1,K_2,K_3$ and $K_4$ respectively
Here,  $C_1=\frac{K_1{\epsilon }_{0}A/2}{d/2}=\frac{K_1{\epsilon }_{0}A}{d}$
Similarity, $C_2=\frac{K_2{\epsilon }_{0}A}{d}{C}_3=\frac{K_3{\epsilon }_{0}A}{d}$ and  $C_4=\frac{K_4{\epsilon }_{0}A}{d}$
Since equivalent capacitance in series combination is
$C_{eq}=\frac{C_1\cdot C_2}{C_1+C_2}$
Here,  and  are in series combination
$\begin{array}{l}\therefore {\left(C_{eq}\right)}_{12}=\frac{C_1\cdot C_2}{C_1+C_2}=\frac{\frac{K_1{\epsilon }_{0}A}{d}\cdot \frac{K_2{\epsilon }_{0}A}{d}}{\frac{K_1{\epsilon }_{0}A}{d}+\frac{K_2{\epsilon }_{0}A}{d}}\\ =\frac{K_1\cdot K_2}{K_1+K_2}\cdot \frac{{\epsilon }_{0}A}{d}\end{array}$
Similarly, ${\left(C_{eq}\right)}_{34}=\frac{K_3{K}_4}{K_3+K_4}\cdot \frac{{\epsilon }_{0}A}{d}$
Now, and  are in parallel combination
$\begin{array}{l}\therefore C_{net}={\left(C_{eq}\right)}_{12}+{\left(C_{eq}\right)}_{34}\\ =\frac{K_1\cdot K_2}{K_1+K_2}\cdot \frac{{\epsilon }_{0}A}{d}+\frac{K_3\cdot K_4}{K_3+K_4}\cdot \frac{{\epsilon }_{0}A}{d}\end{array}$
$\Rightarrow C_{net}=\left(\frac{K_1{K}_2}{K_1+K_2}+\frac{K_3\cdot K_4}{K_3+K_4}\right)\frac{{\epsilon }_{0}A}{d}$…(i)
If K is effective dielectric constant, then
$C_{net}=\frac{K{\epsilon }_{o}A}{d}$
From Eqs (i) and (ii)
$\frac{K{\epsilon }_{0}A}{d}=\left(\frac{K_1\cdot K_2}{K_1+K_2}+\frac{K_3\cdot K_4}{K_3+K_4}\right)\frac{{\epsilon }_{0}A}{d}$
or  $K=\left(\frac{K_1\cdot K_2}{K_1+K_2}+\frac{K_3\cdot K_4}{K_3+K_4}\right)$