A closed surface intersecting a conducting sphere is depicted in...

When a positive charge is placed at point P, it induces a negative charge on the closer side (right side) of the conducting sphere due to electrostatic attraction.

Consequently, an equal amount of positive charge is induced on the farther side (left side) of the conducting sphere due to repulsion.

As seen in the figure, the closed surface encloses the left portion of the conducting sphere. Therefore, the net charge enclosed by the closed surface is positive.

According to Gauss's law, the electric flux through a closed surface is given by $\Phi = \dfrac{q_{\text{enclosed}}}{\epsilon_0}$.

Since the enclosed charge $q_{\text{enclosed}}$ is positive, the electric flux through the closed surface will become positive.