Identify the general form of a progressive wave equation and compare it with the given equation.

The general equation for a progressive wave moving in the positive x-direction is:

$$y = A \sin(\omega t - kx)$$

The given wave equation is:

$$y = a \sin\left(150 \pi t - \frac{\pi x}{4}\right)$$

By comparing these two equations, we can see that the term representing the phase constant due to position $x$ is $\frac{\pi x}{4}$. This term represents the phase difference from the origin at a given position $x$.

The problem states that the phase difference is $45^\circ$. To use this in the equation, we must convert it to radians:

$$\text{Phase difference} = 45^\circ \times \frac{\pi \text{ radians}}{180^\circ}$$ $$\text{Phase difference} = \frac{\pi}{4} \text{ radians}$$

The phase constant in the wave equation is $\frac{\pi x}{4}$. We are given that this phase difference is $\frac{\pi}{4}$ radians.

$$\frac{\pi x}{4} = \frac{\pi}{4}$$

To solve for $x$, we can multiply both sides of the equation by 4:

$$\pi x = \pi$$

Then, divide both sides by $\pi$:

$$x = \frac{\pi}{\pi}$$ $$x = 1$$

The options are given in cm, so the value of $x$ is $1 \text{ cm}$.

The final answer is $\text{D}$.