A slab of dielectric constant K has the same cross-sectional area as the plates of a parallel plate capacitor and thickness $$\frac{3}{4}d$$, where $$d$$ is the separation of the plates. The capacitance of the capacitor when the slab is inserted between the plates will be: (Given $$C_0$$ = capacitance of capacitor with air as medium between plates.)

We have a parallel plate capacitor with plate separation $$d$$ and original capacitance $$C_0 = \frac{\varepsilon_0 A}{d}$$. A dielectric slab of constant $$K$$ and thickness $$\frac{3d}{4}$$ is inserted between the plates.

When a dielectric slab is inserted, the system can be treated as two capacitors in series: one with the dielectric (thickness $$\frac{3d}{4}$$) and one with air (thickness $$d - \frac{3d}{4} = \frac{d}{4}$$).

The capacitance of the dielectric portion is $$C_1 = \frac{K\varepsilon_0 A}{3d/4} = \frac{4K\varepsilon_0 A}{3d}$$.

The capacitance of the air portion is $$C_2 = \frac{\varepsilon_0 A}{d/4} = \frac{4\varepsilon_0 A}{d}$$.

For capacitors in series, $$\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{3d}{4K\varepsilon_0 A} + \frac{d}{4\varepsilon_0 A} = \frac{d}{4\varepsilon_0 A}\left(\frac{3}{K} + 1\right) = \frac{d}{4\varepsilon_0 A} \cdot \frac{3+K}{K}$$.

So $$C = \frac{4K\varepsilon_0 A}{d(3+K)}$$. Since $$C_0 = \frac{\varepsilon_0 A}{d}$$, we can write $$C = \frac{4KC_0}{3+K}$$.

Hence, the correct answer is Option A.