The equivalent capacitance between 'P' and 'Q' is [A-area of each plate, d-separation between the plates, $\epsilon_0$ -permittivity of free space]

The given arrangement is equivalent to three capacitors connected in parallel between the terminals P and Q, as the potential difference across each section is the same.

The capacitance of a parallel plate capacitor with a dielectric is given by $C = \dfrac{K \epsilon_0 A'}{d}$.

For the three sections, the area is $A' = \dfrac{A}{3}$ and the distance between the plates is $d$.

The individual capacitances are:
$C_1 = \dfrac{K_1 \epsilon_0 (A/3)}{d} = \dfrac{1 \cdot \epsilon_0 A}{3d} = \dfrac{\epsilon_0 A}{3d}$

$C_2 = \dfrac{K_2 \epsilon_0 (A/3)}{d} = \dfrac{2 \cdot \epsilon_0 A}{3d} = \dfrac{2\epsilon_0 A}{3d}$

$C_3 = \dfrac{K_3 \epsilon_0 (A/3)}{d} = \dfrac{3 \cdot \epsilon_0 A}{3d} = \dfrac{3\epsilon_0 A}{3d}$

Since the capacitors are in parallel, the equivalent capacitance $C_{eq}$ is the sum of the individual capacitances:

$C_{eq} = C_1 + C_2 + C_3$

$C_{eq} = \dfrac{\epsilon_0 A}{3d} + \dfrac{2\epsilon_0 A}{3d} + \dfrac{3\epsilon_0 A}{3d}$

$C_{eq} = \dfrac{\epsilon_0 A}{3d} (1 + 2 + 3)$

$C_{eq} = \dfrac{6\epsilon_0 A}{3d} = \dfrac{2A\epsilon_0}{d}$

Answer: $\dfrac{2A\epsilon_0}{d}$