1. Principle to use

For a charged conductor in electrostatic equilibrium:

• Charge resides on the outer surface.
• The surface charge density is greater where the surface is more sharply curved (smaller radius of curvature).
• At geometrically equivalent/symmetric points, charge density is equal.

So, to compare $\sigma_1,\sigma_2,\sigma_3,\sigma_4$, we use the local curvature of the irregular metallic disk.

2. Comparing the marked points

From the figure (irregular metallic disk), point $1$ is on the sharpest protruding region, point $3$ is on a less sharp curved region, and points $2$ and $4$ are on similar broader regions.

Hence: $\sigma_1 > \sigma_3 > \sigma_2 = \sigma_4$

3. Check each statement

• Statement A: $\sigma_1 > \sigma_3$ and $\sigma_2 = \sigma_4$

  This is true.

• Statement B: $\sigma_1 > \sigma_2$ and $\sigma_3 > \sigma_4$

  Since $\sigma_1 > \sigma_3 > \sigma_2 = \sigma_4$, both parts are true. So B is true.

• Statement C: $\sigma_1 > \sigma_3 > \sigma_2 = \sigma_4$

  This matches the correct ordering exactly, so C is true.

• Statement D: $\sigma_1 < \sigma_3 < \sigma_2 = \sigma_4$

  Opposite of the actual trend, so false.

• Statement E: $\sigma_1 = \sigma_2 = \sigma_3 = \sigma_4$

  This would be true only for a highly symmetric shape like a sphere, not for an irregular conductor. So false.

4. Selecting the correct option

True statements are A, B and C only.

Therefore, the correct option is: $\boxed{D}$

5. Comparison with stored answer

Stored correct answer: $D$

Our derived answer also is $D$, so they agree.