You're not alone in searching a pot with 90 ° rotation angle. All references I can find mention standard 270 ° pots with mechanically limited rotation angle. The ones that have "90 °" in the title of the sales offer refer to the angle between the pins and the potentiometer axle as 90 °, not the rotation angle.
In your circuit the pot is used as a voltage divider together with the 10 kΩ resistor at the lower end and the 100 kΩ resistor at the upper end. It thus delivers these voltages to pin 3 of IC1a:
0.05 × V+ when in the lower position,
0.52 × V+ when in the upper position
You cannot achieve this exact ratio by simple means. You can approximate it, however:
- Remove the upper 100 kΩ resistor by a short circuit.
- Change the lower 10 kΩ resistor to 5 kΩ (use 2 × 10 kΩ in parallel or use 4.7 kΩ, good enough)
- Change the gain of IC1a from 1×to 2× by adding two resistors in the feedback as shown e.g. here. The value of the 2 additional resistors is not important. both shall have the same vale in the range 1 kΩ to 10 kΩ, whatever you have in your parts bin.
- Mechanically limit the potentiometer to move from the left side (lower pin in the diagram) to left side + 90 °.
Using 4.7 k will give you these divider ratios:
0.1 × V+ in the lower position
0.72 × V+ in the upper position.
You can tweak (fine tune) these values by changing the gain f the opamp IC1a from 2 to another (lower) value. Do this by changing the resistor values in the feedback circuit according to the equations given in the link I providd above.