1. Use energy relation in SHM

For a particle in SHM with amplitude $a$, the total energy is

$E = \frac{1}{2}k a^2$

At displacement $y$, the potential energy is

$U = \frac{1}{2}k y^2$

So the kinetic energy is

$K = E - U = \frac{1}{2}k(a^2 - y^2)$

Using $E = \frac{1}{2}k a^2$, we can write

$K = E\left(1 - \frac{y^2}{a^2}\right)$

2. Given kinetic energy

It is given that

$K = \frac{3E}{4}$

So,

$E\left(1 - \frac{y^2}{a^2}\right) = \frac{3E}{4}$

Since $E \neq 0$, divide both sides by $E$:

$1 - \frac{y^2}{a^2} = \frac{3}{4}$

$\frac{y^2}{a^2} = 1 - \frac{3}{4} = \frac{1}{4}$

$y^2 = \frac{a^2}{4}$

$y = \pm \frac{a}{2}$

3. Choose from the options

Since the options list only the magnitude of displacement, we take

$y = \frac{a}{2}$

So the correct option is D.