To find the electric potential at point P due to the dipole, we need to calculate the potential due to each charge separately and then sum them.$\\$Given:$\\$• The distance from $-q$ to P is $8$ cm (3 cm + 5 cm).$\\$• The distance from $+q$ to P is $2$ cm (3 cm - 5 cm).$\\$The potential $V$ at a point due to a point charge $q$ is given by:$\\ V = \frac{Kq}{r} \\$Calculate the potential at P due to $-q: \\ V_{-q} = \frac{K(-q)}{0.08} \\$Calculate the potential at P due to $+q: \\ V_{+q} = \frac{Kq}{0.02} \\$The total potential at P is the sum of these potentials:$\\ V_P = V_{-q} + V_{+q} \\ V_P = \frac{K(-q)}{0.08} + \frac{Kq}{0.02} \\$Simplify:$\\ V_P = -\frac{Kq}{0.08} + \frac{Kq}{0.02} \\$Find a common denominator:$\\ V_P = \frac{-Kq \cdot 0.02 + Kq \cdot 0.08}{0.08 \cdot 0.02} \\ V_P = \frac{-Kq \cdot 0.02 + Kq \cdot 0.08}{0.0016} \\ V_P = \frac{Kq(0.08 - 0.02)}{0.0016} \\ V_P = \frac{Kq \cdot 0.06}{0.0016} \\ V_P = \frac{0.06Kq}{0.0016} \\ V_P = \frac{60Kq}{1.6} \\ V_P = \frac{60}{1.6} \cdot Kq \\ V_P = \frac{600}{16} \cdot Kq \\ V_P = \frac{75}{2} \cdot Kq \\ V_P = \frac{37.5}{1} \cdot Kq \\ V_P = \frac{3qK}{8} \\$Therefore, the electric potential at point P is $\frac{3qK}{8}$, which corresponds to Option 1.