This is the thread with the original question.

The line PB is

*not* at a 45 degree angle with the line PQ. If F is the projection of the balloon on the ground, then 45 degrees is between PB and PF. In any case, the 60 degrees bearing angle is on the ground, while the angle between PB and PQ is in a different plane.

Follow the suggestion of earboth in the thread above and find QF, FB and PF. Show that angle PFQ is 90 degrees. Knowing the sides of triangle PFQ, you can find angle FPQ. Then the required bearing is 60 + that angle.

Hint: it is not a coincidence that in my picture Q lies on the y axis.