Your reasoning is correct, but you need to go a little further to deduce the answer.

The question does not say but we must assume that the mean position O is the same for P and Q. The phase difference between them remains the same, but the separation changes.

Suppose that, at some instant, P and Q are moving in the same direction. When the separation is maximum they are moving with the same speed. If the speed of one was greater than the other, they would get closer together of further apart. For each, speed decreases with distance from O in the same manner, because they have the same amplitude. Whichever is closer to O moves faster. They will have the same speed when they are the same distance from O. So at maximum separation they will be positioned symmetrically about O.

If the maximum separation is $A\sqrt2$ then each is $\frac{A\sqrt2}{2}=\frac{A}{\sqrt2}$ from O. The phase of each is then $\phi$ where

$A\sin\phi=\frac{A}{\sqrt2}$

$\sin\phi=\frac{1}{\sqrt2}=\sin45^{\circ}$

$\phi=45^{\circ}$.

So the phase difference at maximum separation (and at all separations) is $2\phi=90^{\circ}$.

Alternative solution :

$x_1=A\sin(\omega t+\phi)$

$x_2=A\sin(\omega t)$

$x_1-x_2=2A\cos(\frac{\omega t+\phi+\omega t}{2})\sin(\frac{\omega t+\phi - \omega t}{2})=2A\cos(\omega t+\frac12 \phi)\sin(\frac12 \phi)$.

$\sin(\frac12 \phi)$ is a constant, so the maximum possible value of $x_1-x_2$ occurs when $\cos(\omega t+\frac12 \phi)=1$. Then

$2A\sin(\frac12 \phi)= A\sqrt2$

$\sin(\frac12 \phi)=\frac{\sqrt2}{2}=\frac{1}{\sqrt2}=\sin 45^{\circ}$

$\phi=90^{\circ}$.