So big picture!!! Please resize it.
From the given information, we can calculate the radius of the circle : r^2 = x^2 + y^2 = (-sqrt(3)) ^2 + 1^2 = 4, then r = 2
As OP and OQ crosses the original O, the equation of OP and OQ lines must be in the form: y = a*x
Plug in the given point P(-sqrt(3), 1) into OP's equation: 1 = a * -sqrt(3), or a = 1/-sqrt(3)
As line OP is perpendicular to line OQ, a1 ( a of OP's equation) * a2 ( a of OQ's equation) = -1
a2 = -1/ a1 = -1/ (1/-sqrt(3)) = sqrt(3)
the equation of line OQ: y = sqrt(3)*x, plug in Q(s,t) into OQ's equation: t = sqrt(3) * s or t^2 = 3 * s^2
Because Q (s,t) is on the circle, s^2 + t^2 = r^2 or s^2 + t^2 = 2^2
From the two bold equations : 4* s^2 = 4
Or s^2 = 1 or s = +1 or -1
However, Q is to the right the Oy vector, hence s must be equal to 1.
B is the answer.

Quick approach:

Slope of OP = rise/run = (1)/ (-√3).
To get from the origin to point P (the radius), we need to move up 1 and to the left √3.
OQ is perpendicular to OP because the two lines form a right angle. The slopes of perpendicular lines are negative reciprocals.
Thus the slope of OQ = √3/1.
OQ = radius, so point Q must be the same distance from the origin as point P.
Thus, to get from the origin to point Q, we need to move up √3 and to the right 1. This is the only way to have a slope of √3/1 and be the same distance from the origin as point P.
So s=1.

Second approach:
When you see √3 in a problem, look for a 30-60-90 triangle.
The attached .pdf shows how you could draw a line from P to the x axis and another from Q to the x axis to create two 30-60-90 triangles and determine the value of s.

Good luck on your GMAT!