Here’s my go (click to embiggen)

Steps:

Connect A to P and pick an arbitrary point Q between them, near-ish to P.

Then, draw lines as shown, constructing the points in alphabetical order, which correspond to colours in the brightening order of black-red-blue-green.

Notice in particular how each new point is at a clearly defined intersection, and that we’re only using the straightedge to draw lines through two previously constructed points, just as we should.

Once we’re done, point V will be opposite P, and can be used to draw the perpendicular. (Dotted line)

The key points of why this works are

1. The angles ARB and APB are both right angles, because they are inscribed angles subtended by the circle diameter.

2. Therefore, Q is the orthocentre of triangle ABS: the right angles mean that AP and BR are both altitudes of triangle ABS, and all altitudes of a triangle meet at the orthocentre.

3. Therefore, the line through S and Q (blue) must also be an altitude of triangle ABS, and thus, ST is perpendicular to AB.

4. Using the diameter AB and the newly found perpendicular ST, it is now easy to mirror P to V. (green)

To reconfirm that this works for all points on the circle, we only need to notice that the horizontal position of the perpendicular ST doesn’t actually matter; we can use this same ST for any other point P’ on the circle, and we only need to construct two new points (U’ and V’) in order to find the new perpendicular at P’.

EDIT: In the comments, @FlorianF suggests I may have missed a more direct approach by drawing a line through R and the intersection point of AB and ST, and what do you know: