When lead sulphide (PbS) reacts with dilute nitric acid, the reaction is a redox process where lead sulphide is oxidized and nitric acid is reduced. The overall reaction involves the transformation of PbS and HNO₃ to form lead nitrate (Pb(NO₃)₂), sulphur, and nitrogen oxides, along with water. The specific nitrogen oxide formed depends on the reaction conditions, including the concentration of nitric acid and the temperature. Typically, with dilute nitric acid, nitric oxide (NO) and nitrogen dioxide (NO₂) are more common products than nitrous oxide (N₂O).

The general reaction can be represented as:

$$\text{PbS} + 4\text{HNO}_3 \rightarrow \text{Pb(NO}_3\text{)}_2 + \text{SO}_2 + 2\text{H}_2\text{O} + 2\text{NO}$$

Or, under some conditions, sulphur (S) may also be one of the products instead of or in addition to SO₂:

$$\text{PbS} + 2\text{HNO}_3 \rightarrow \text{Pb(NO}_3\text{)}_2 + \text{S} + 2\text{H}_2\text{O}$$

From the reaction above, you can see:

• Sulphur (Option A) can be formed depending on the specific conditions of the reaction, making it a potentially correct product.
• Nitric oxide (Option B) is typically produced when dilute nitric acid reacts with PbS, making it a correct product.
• Lead nitrate (Option C) is definitely formed in the reaction, making it a correct product.
• However, Nitrous oxide (Option D) is not a typical product of the reaction of lead sulphide with dilute nitric acid. It could be formed under special circumstances in reactions involving nitrogen compounds, but it's not a standard outcome for the conditions described.

Therefore, the correct answer to the question is Option D: Nitrous oxide. It is not a typical product of the reaction of lead sulphide with dilute nitric acid under normal conditions.