As we know that, the required formula of Potential energy executing simple harming motion is as below,

$P. E. = \frac{1}{2}kx^{2} And we also know that, \omega^{2} = \frac{k}{m} As from the above equation we need the value of k from the above equation, after getting the value we use it in Potential energy's formula to get the result, k = \omega^{2}m Putting the value we get the result as, P. E. = \frac{1}{2}m\omega^{2}x^{2} As a result, we get the Potential energy of the particle executing simple harmonic motion from the distance x from the equilibrium position is \frac{1}{2}m\omega^{2}x^{2}. Therefore, the correct answer is \frac{1}{2}m\omega^{2}x^{2}.$