Each side of the dodecagon subtends an angle of $30^{\circ}$ at the
centre of the circumcircle of the figure (the circle which passes
through all $12$ of its vertices).
$$\angle AOP = 90^{\circ}$$ so $$\angle OPA = 45^{\circ}$$
$$\angle BOP = 120^{\circ}$$ so $$\angle OPB = 30^{\circ}$$
Therefore $$\angle APB = 45^{\circ} + 30^{\circ} = 75^{\circ}$$
Alternatively, $$\angle AOB = 150^{\circ}$$ and, as the angle
subtended by an arc at the centre of a circle is twice the angle
subtended by that arc at a point on the circumference, $$\angle APB
= 75^{\circ}$$