Two particles 'A' and 'B' execute SHMs of periods 'T' and $'\left.\frac{3 T}{2}\right.'$. If they start from the mean position then the phase difference between them, when the particle 'A' completes two oscillations will be

Given the periods $T_A = T$ and $T_B = \frac{3T}{2}$, both particles begin at the mean position with initial phase zero.

Particle A completes two oscillations after $t = 2T_A = 2T$ seconds.

The angular frequencies are $\omega_A = \frac{2\pi}{T_A} = \frac{2\pi}{T}$ and $\omega_B = \frac{2\pi}{T_B} = \frac{4\pi}{3T}$.

At $t = 2T$, their phases are:

$\phi_A = \omega_A t = \frac{2\pi}{T} \cdot 2T = 4\pi$

$\phi_B = \omega_B t = \frac{4\pi}{3T} \cdot 2T = \frac{8\pi}{3}$

The phase difference is:

$\Delta\phi = \left|4\pi - \frac{8\pi}{3}\right| = \frac{4\pi}{3}$

Final answer: $\boxed{\text{C}}$