To understand why the third dark band occurs at point P in Young's double slit interference experiment, let's first go over the basics of the experiment and the conditions for bright and dark bands.

1. Young's Double Slit Experiment:
In Young's double slit interference experiment, a beam of light passes through two narrow slits and forms an interference pattern on a screen placed behind the slits. This pattern is a result of the superposition of two coherent waves generated by the slits.

2. Conditions for Bright and Dark Bands:
The interference pattern consists of alternating bright and dark bands. The bright bands occur where the waves are in phase and undergo constructive interference, resulting in a bright region on the screen. The dark bands occur where the waves are out of phase and undergo destructive interference, resulting in a dark region on the screen.

3. Path Difference:
Path difference is the difference in the distances traveled by the two waves from the two slits to a specific point on the screen. It determines whether constructive or destructive interference occurs at that point.

4. Relationship between Path Difference and Bright/Dark Bands:
The path difference between the two waves determines the order of the bright and dark bands. For constructive interference (bright bands), the path difference should be an integer multiple of the wavelength (λ) of the light used. For destructive interference (dark bands), the path difference should be a half-integer multiple of the wavelength.

Now, let's apply these concepts to the given scenario:

- The wavelength of the light used is 6000 Å (angstroms), which is equivalent to 6000 x 10^-10 meters or 6000 x 10^-7 microns.
- The path difference between the waves reaching point P on the screen is 1.5 microns, which is 1.5 x 10^-6 meters or 1.5 x 10^-3 microns.

To determine whether a bright or dark band occurs at point P, we need to compare the path difference with the wavelength:

- For constructive interference (bright bands), the path difference should be an integer multiple of the wavelength. In this case, 1.5 microns is not an integer multiple of 6000 Å, so a bright band does not occur.
- For destructive interference (dark bands), the path difference should be a half-integer multiple of the wavelength. In this case, 1.5 microns is equal to 1.5 x 10^-3 microns, which is half of 3 x 10^-3 microns (which is 6000 Å). Therefore, a dark band occurs at point P.

Hence, the correct answer is option 'C' - Third dark band occurs.