Consider the given molecule with four labeled C-Cl bonds (P, Q, R,...

The ease of heterolytic cleavage of a $\text{C}-\text{Cl}$ bond depends on the stability of the carbocation formed after the departure of the chloride ion ($\text{Cl}^-$). The more stable the carbocation, the more readily the bond will cleave.

Let us analyze the carbocations formed by the cleavage of each labeled bond:

1. Bond P: Cleavage forms a tertiary benzylic carbocation. This carbocation is highly stabilized by resonance with the benzene ring and by hyperconjugation from the six $\alpha$-hydrogens of the two methyl groups.

2. Bond Q: Cleavage forms a primary benzylic carbocation. It is stabilized by resonance with the benzene ring, but it is less stable than the tertiary benzylic carbocation formed from P.

3. Bond R: Cleavage forms a phenyl cation. This is highly unstable because the positive charge resides in an $sp^2$ hybridized orbital that is perpendicular to the $\pi$-system of the benzene ring, meaning it cannot be stabilized by resonance. Furthermore, the $\text{C}-\text{Cl}$ bond here has partial double bond character due to resonance, making it difficult to break.

4. Bond S: Cleavage forms a primary alkyl carbocation ($\text{-CH}_2\text{-CH}_2^+$). This carbocation lacks resonance stabilization and is relatively unstable.

Comparing the stabilities of the resulting carbocations, the order is:
Tertiary benzylic (from P) > Primary benzylic (from Q) > Primary alkyl (from S) > Phenyl (from R).

Since the carbocation formed from the cleavage of bond P is the most stable, bond P will undergo heterolytic cleavage most readily.

Answer: Bond P