If $$q_f$$ is the free charge on the capacitor plates and $$q_b$$ is the bound charge on the dielectric slab of dielectric constant $$k$$ placed between the capacitor plates, then bound charge $$q_b$$ can be expressed as:

We begin by recalling the fundamental relation that connects the electric displacement vector $$\mathbf D$$ with the free surface charge density $$\sigma_f$$ on a dielectric-filled capacitor plate. By definition,

$$\mathbf D \cdot \hat n \;=\; \sigma_f.$$

Here $$\hat n$$ is the outward normal to the surface and $$\sigma_f$$ is linked to the total free charge $$q_f$$ on the plate of area $$A$$ through

$$\sigma_f \;=\; \dfrac{q_f}{A}.$$

Next, we write the constitutive relation between $$\mathbf D$$, the electric field $$\mathbf E$$ and the polarisation $$\mathbf P$$ inside a linear isotropic dielectric:

$$\mathbf D \;=\; \varepsilon_0 \mathbf E \;+\; \mathbf P.$$

The dielectric constant $$k$$ (relative permittivity $$\varepsilon_r$$) is defined by

$$k = \varepsilon_r \;=\; 1 \;+\; \chi_e,$$

where $$\chi_e$$ is the electric susceptibility. Therefore

$$\mathbf P \;=\; \varepsilon_0 \chi_e \mathbf E \;=\; \varepsilon_0 (k-1)\mathbf E.$$

The bound surface charge density $$\sigma_b$$ appears wherever the polarisation vector $$\mathbf P$$ terminates on a surface, and it is given by the formula

$$\sigma_b \;=\; \mathbf P \cdot \hat n.$$

Substituting the expression of $$\mathbf P$$ we obtain

$$\sigma_b \;=\; \varepsilon_0 (k-1) \mathbf E \cdot \hat n \;=\; \varepsilon_0 (k-1) E,$$

because $$\mathbf E$$ is perpendicular to the plate and points along $$\hat n$$ in a parallel-plate capacitor.

In precisely the same geometry the displacement vector is purely normal, giving

$$\sigma_f \;=\; \mathbf D \cdot \hat n \;=\; \varepsilon_0 k E.$$

Now we take the ratio of the bound to the free surface charge densities:

$$\dfrac{\sigma_b}{\sigma_f} \;=\; \dfrac{\varepsilon_0 (k-1) E}{\varepsilon_0 k E} \;=\; \dfrac{k-1}{k} \;=\; 1 - \dfrac{1}{k}.$$

Since both surface charge densities refer to the same plate area $$A$$, multiplying by $$A$$ converts them to the corresponding total charges:

$$q_b \;=\; \sigma_b A, \qquad q_f \;=\; \sigma_f A.$$

Therefore the ratio of the total charges is identical to the ratio of the surface charge densities:

$$\dfrac{q_b}{q_f} \;=\; 1 - \dfrac{1}{k}.$$

Solving for the bound charge $$q_b$$ we finally obtain

$$q_b \;=\; q_f\left(1 - \dfrac{1}{k}\right).$$

Hence, the correct answer is Option B.