A particle is executing S.H.M. of amplitude 'A'. When the potential energy of the particle is half of its maximum value during the oscillation, its displacement from the equilibrium position is

The potential energy at displacement $x$ from equilibrium is $PE = \frac{1}{2}kx^2$, where $k$ is the force constant. Maximum potential energy occurs at amplitude $A$, giving $PE_{\text{max}} = \frac{1}{2}kA^2$.

Given that $PE = \frac{1}{2}PE_{\text{max}}$, substitution yields:

$\frac{1}{2}kx^2 = \frac{1}{2}\left(\frac{1}{2}kA^2\right)$

Simplifying both sides gives $\frac{1}{2}kx^2 = \frac{1}{4}kA^2$. Canceling the common factor $\frac{1}{2}k$ produces $x^2 = \frac{1}{2}A^2$, so $x = \pm \frac{A}{\sqrt{2}}$.

The displacement when potential energy is half its maximum is $\boxed{\text{D}}$.