The space between the plates of a parallel plate capacitor is filled with a 'dielectric' whose 'dielectric constant' varies with distance as per the relation: $$K(x) = K_o + \lambda x$$ ($$\lambda$$ = a constant). The capacitance C, of the capacitor, would be related to its vacuum capacitance C$$_o$$ for the relation:

The capacitance of a parallel plate capacitor with a varying dielectric constant can be found by considering the capacitor as composed of infinitesimal capacitors in series. The dielectric constant is given by $$K(x) = K_o + \lambda x$$, where $$x$$ is the distance from one plate, and the plates are separated by a distance $$d$$. The vacuum capacitance is $$C_o = \frac{\varepsilon_o A}{d}$$, where $$A$$ is the plate area and $$\varepsilon_o$$ is the permittivity of free space.

For an infinitesimal capacitor of thickness $$dx$$ at position $$x$$, the capacitance is $$dC = \frac{K(x) \varepsilon_o A}{dx}$$. Since these infinitesimal capacitors are in series, the reciprocal of the total capacitance $$C$$ is the integral of the reciprocals of the infinitesimal capacitances:

$$\frac{1}{C} = \int_0^d \frac{1}{dC} dx = \int_0^d \frac{dx}{K(x) \varepsilon_o A}$$

Substituting $$K(x) = K_o + \lambda x$$:

$$\frac{1}{C} = \int_0^d \frac{dx}{(K_o + \lambda x) \varepsilon_o A} = \frac{1}{\varepsilon_o A} \int_0^d \frac{dx}{K_o + \lambda x}$$

To solve the integral, use the substitution $$u = K_o + \lambda x$$. Then $$du = \lambda dx$$, so $$dx = \frac{du}{\lambda}$$. The limits change: when $$x = 0$$, $$u = K_o$$; when $$x = d$$, $$u = K_o + \lambda d$$. The integral becomes:

$$\int_0^d \frac{dx}{K_o + \lambda x} = \int_{K_o}^{K_o + \lambda d} \frac{1}{u} \cdot \frac{du}{\lambda} = \frac{1}{\lambda} \int_{K_o}^{K_o + \lambda d} \frac{du}{u} = \frac{1}{\lambda} \left[ \ln|u| \right]_{K_o}^{K_o + \lambda d} = \frac{1}{\lambda} \left( \ln(K_o + \lambda d) - \ln(K_o) \right) = \frac{1}{\lambda} \ln\left(\frac{K_o + \lambda d}{K_o}\right) = \frac{1}{\lambda} \ln\left(1 + \frac{\lambda d}{K_o}\right)$$

Substituting back:

$$\frac{1}{C} = \frac{1}{\varepsilon_o A} \cdot \frac{1}{\lambda} \ln\left(1 + \frac{\lambda d}{K_o}\right)$$

Recall that $$C_o = \frac{\varepsilon_o A}{d}$$, so $$\varepsilon_o A = C_o d$$. Substituting this in:

$$\frac{1}{C} = \frac{1}{C_o d} \cdot \frac{1}{\lambda} \ln\left(1 + \frac{\lambda d}{K_o}\right) = \frac{\ln\left(1 + \frac{\lambda d}{K_o}\right)}{\lambda C_o d}$$

Solving for $$C$$:

$$C = \frac{\lambda C_o d}{\ln\left(1 + \frac{\lambda d}{K_o}\right)} = \frac{\lambda d}{\ln\left(1 + \frac{\lambda d}{K_o}\right)} C_o$$

Comparing with the options, this matches Option C.

Hence, the correct answer is Option C.