The total number of sp$$^2$$ hybridised carbon atoms in the major product P (a non-heterocyclic compound) of the following reaction is ______.

Step 1: Reduction with $$\text{LiAlH}_4$$ (excess)

The starting material is $$1,1,2,2\text{-tetracyanoethane}$$:

$$\text{(NC)}_2\text{CH}-\text{CH(CN)}_2$$

Lithium aluminum hydride ($$\text{LiAlH}_4$$) is a strong reducing agent that reduces all four nitrile ($$-\text{C}\equiv\text{N}$$) groups completely into primary amine ($$-\text{CH}_2\text{NH}_2$$) groups.

The intermediate tetra-amine formed after workup with $$\text{H}_2\text{O}$$ is:

$$\text{(H}_2\text{NCH}_2)_2\text{CH}-\text{CH}(\text{CH}_2\text{NH}_2)_2$$

Step 2: Reaction with Acetophenone (excess)

Acetophenone ($$\text{Ph}-\text{CO}-\text{CH}_3$$) reacts with primary amines ($$-\text{NH}_2$$) via a nucleophilic addition-elimination reaction to form imines ($$-\text{N}=\text{C}\langle$$):

$$\text{R}-\text{NH}_2 + \text{O}=\text{C}(\text{CH}_3)\text{Ph} \longrightarrow \text{R}-\text{N}=\text{C}(\text{CH}_3)\text{Ph} + \text{H}_2\text{O}$$

Since acetophenone is in excess, all four primary amine groups will react to form four imine groups in the final product $$P$$.

Step 3: Counting $$\text{sp}^2$$ Hybridised Carbon Atoms

Let's analyze the hybridization of carbons in one attached acetophenone-derived unit ($$-\text{N}=\text{C}(\text{CH}_3)\text{Ph}$$):

1. Imine Carbon ($$-\text{N}=\underline{\text{C}}-$$): $$1\text{ sp}^2$$ carbon
2. Methyl Carbon ($$-\text{CH}_3$$): $$0\text{ sp}^2$$ carbons ($$\text{sp}^3$$)
3. Phenyl Ring ($$-\text{C}_6\text{H}_5$$): All $$6$$ carbons in the benzene ring are $$\text{sp}^2$$ hybridised.

$$\text{Number of sp}^2\text{ carbons per unit} = 1 + 6 = 7$$

$$\text{Total number of sp}^2\text{ hybridised carbons} = 4 \times 7 = \mathbf{28}$$