As a retired engineer I am going to supply the solution alike one of my draftmen will probably would have given to me.
1) Chose any diameter D in your circunference c and center O draw an isosceles triangle, being D the base and others two equal sides, forming the given angle, let us call it W
2) Draw a new circunference q passing for the three vertexs of the isosceles triangle. So any given point of the circunference will form an angle W with the ends points of the D which has became a chord of circunference q.
3) Draw another circunference r with the same center O that the original circunference c and passing for the given point P. The circunference r the radius OP will intersect the circunference q in two, zero points an exceptionally in one.
4) If there is zero point there is no real solution, if there are two point lets call them M1 and M2 there are two solutions. Lets take one of the two points, M1 draw a radius OM1, which will form and angle X with OP.
5 ) Now all you have to do it with center in O rotate the circunference q including the diameter D (a chord of circunference q.) an angle X, so M1 will transform into P, and the original diameter D will rotated to D1 which is one of the solutions. The same procedure with the other M2.
Of course you can attack the problem using vectors, is an elegant concise equation using the scalar product. However to solve the radical that will appears in the denominator ends in a cumbersome quartic equation with two or four non real solutions.
Same things will happen using complex numbers
There is at least one more beautiful approach using stereographic projection somewhat linked to AlexB response but for me with English as a second language it is very difficult to express. But I believe the above one is real simple.