This question tests two concepts. Properties of a circle and special right angled triangles.
Refer to the image attached:

In triangle PAO, coorinates of P are (-sqrt(3),1). This means that PA is sqrt(3) and OA is 1.
Therefore PAO is a 30-60-90 triangle where AO is 1, PO is sqrt(3) and PO is 2.

Now since PO = OQ = radius of circle = 2, we can find what the co-ordinates of Q are.

We are shown in the question that angle POQ = 90, since we already found that angle POB =60, therefore angle BOQ = 90-60 = 30.

Now triangle BOQ is another 30-60-90 triangle. Since OQ = 2, and BQ is opposite 30 degree angle, BQ = 1. So co-ordinates of Q are (1,sqrt(3). Hence s=1, t= sqrt(3).

Let me know if this helps

Anurag@Gurome: Your solution was fine but it didn't work for me. The 30-60-90 rule suggested by eagleeye is something that would click first to me on the real test. Thanks, nevertheless.