Answer
The light intensity $I_p$ is multiplied by $~~1.11~~$ by the presence of the mirror.
Work Step by Step
After the mirror is placed behind the source of light, there are two sources of light on the screen. The mirror adds a second source of light that we can say is a distance of $3d$ from the screen.
Since the intensity of light is inversely proportional to the square of the distance, the intensity at the screen is increased by $\frac{I_p}{9}$ which is $0.11~I_p$
Therefore, the light intensity $I_p$ is multiplied by $1.11$ by the presence of the mirror.