Intensity after the first two polarizers

The incident light is unpolarized with intensity $I_0$.

When unpolarized light passes through a polarizer, its intensity becomes half: $I_1 = I_2 = \frac{I_0}{2}$

So, after $P_1$ and $P_2$, the two beams emerging from slits $s_1$ and $s_2$ each have intensity $\frac{I_0}{2}$.

Effect of the third polarizer $P_3$

The transmission axes of $P_1$ and $P_2$ are mutually perpendicular, and $P_3$ is at $45^\circ$ to both.

Each beam passes through $P_3$ at an angle $45^\circ$ with its polarization direction. By Malus' law: $I' = I \cos^2 45^\circ = I \cdot \frac{1}{2}$

Therefore, intensity of each beam after $P_3$ is $I_1' = I_2' = \frac{I_0}{2}\cdot \frac{1}{2} = \frac{I_0}{4}$

Now both beams have the same polarization direction (along the axis of $P_3$), so they can interfere.

Phase difference corresponding to path difference $\lambda/3$

Given path difference: $\Delta x = \frac{\lambda}{3}$

Hence phase difference is $\phi = \frac{2\pi}{\lambda}\cdot \frac{\lambda}{3} = \frac{2\pi}{3}$

Resultant intensity due to interference

For two coherent waves of intensities $I_1'$ and $I_2'$: $I = I_1' + I_2' + 2\sqrt{I_1'I_2'}\cos\phi$

Since $I_1' = I_2' = \frac{I_0}{4}$, $I = \frac{I_0}{4} + \frac{I_0}{4} + 2\sqrt{\frac{I_0}{4}\cdot \frac{I_0}{4}}\cos\frac{2\pi}{3}$

$I = \frac{I_0}{2} + 2\cdot \frac{I_0}{4}\cdot \left(-\frac{1}{2}\right)$

$I = \frac{I_0}{2} - \frac{I_0}{4} = \frac{I_0}{4}$

Correct option

$\boxed{I = \frac{I_0}{4}}$

So the correct option is C.