1. Interpret the figure

   The standard arrangement consistent with the options is:

   • two charges $+q$ and $-q$ separated by distance $d$,
   • a third charge $Q$ placed at a distance $D$ from the midpoint of the dipole,
   • with $D \gg d$.

   So the total electrostatic potential energy is the sum of pairwise interaction energies.

2. Write total potential energy

   For three charges, $U = \frac{1}{4\pi\varepsilon_0}\left( \frac{q_1q_2}{r_{12}} + \frac{q_1q_3}{r_{13}} + \frac{q_2q_3}{r_{23}} \right).$

   Here:

   • interaction between $+q$ and $-q$ gives $U_{q,-q} = \frac{1}{4\pi\varepsilon_0}\left(\frac{(+q)(-q)}{d}\right)= -\frac{1}{4\pi\varepsilon_0}\frac{q^2}{d}.$

3. Energy due to charge $Q$ with the dipole charges

   Let distances of $Q$ from $+q$ and $-q$ be approximately: $r_+ = D - \frac d2, \qquad r_- = D + \frac d2$ (or vice versa; the final correction depends on which side $+q$ lies, and from the options we need the sign corresponding to the shown figure).

   Then

   = \frac{1}{4\pi\varepsilon_0}qQ\left(\frac{1}{r_+}-\frac{1}{r_-}\right).$$

4. Use binomial approximation for $D \gg d$

   Using $\frac{1}{D\pm d/2} \approx \frac{1}{D} \mp \frac{d}{2D^2},$ we get

   \approx \left(\frac{1}{D}+\frac{d}{2D^2}\right)-\left(\frac{1}{D}-\frac{d}{2D^2}\right) = \frac{d}{D^2}.$$ Depending on which charge is nearer to $Q$, the correction is $\pm \dfrac{qQd}{D^2}$. From the given answer options and the figure’s implied orientation, the contribution is $$U_Q = -\frac{1}{4\pi\varepsilon_0}\frac{qQd}{D^2}.$$

5. Add both parts

   Therefore, $U = -\frac{1}{4\pi\varepsilon_0}\frac{q^2}{d} - \frac{1}{4\pi\varepsilon_0}\frac{qQd}{D^2}.$

   Hence, $U=\frac{1}{4\pi\varepsilon_0}\left(-\frac{q^2}{d}-\frac{qQd}{D^2}\right).$

6. Match with options

   This matches Option C.

Final Answer: $\boxed{\text{C}}$