Let $\angle ABC = \angle ACB = x$. Then trivial angle chasing yields - $\angle ACP = x - 30^{\circ}$ $\angle PBC = 111^{\circ} - x$ $\angle ABP = 2x - 111^{\circ}$ $\angle BAP = 141^{\circ} - 2x$ Now applying $\text{trigonometric form of Ceva's Theorem}$ in $\triangle ABC$ yields - $\frac{sin(141^{\circ} - 2x)}{sin39^{\circ}}.\frac{sin(x - 30^{\circ})}{sin30^{\circ}}.\frac{sin(111^{\circ} - x)}{sin(2x - 111^{\circ})} = 1$ $\implies 2sin(141^{\circ} - 2x)sin(x - 30^{\circ})sin(111^{\circ} - x) = sin39^{\circ}sin(2x - 111^{\circ})$ $\implies sin(141^{\circ} - 2x)[cos(2x - 141^{\circ}) - cos81^{\circ}] = sin39^{\circ}sin(2x - 111^{\circ})$ Note that $cos(-\alpha) = cos\alpha$. So, $\implies sin(141^{\circ} - 2x)cos(141^{\circ} - 2x) - sin(141^{\circ} - 2x)cos81^{\circ} = sin39^{\circ}sin(2x - 111^{\circ})$ $\implies 2sin(141^{\circ} - 2x)cos(141^{\circ} - 2x) - 2sin(141^{\circ} - 2x)cos81^{\circ} = 2sin39^{\circ}sin(2x - 111^{\circ})$ $\implies sin(282^{\circ} - 4x) - sin(222^{\circ} - 2x) - sin(60^{\circ} -2x) = cos(150^{\circ} - 2x) - cos(2x - 72^{\circ})$ Again, $sin(60^{\circ} -2x) = -cos(150^{\circ} - 2x)$ yields - $sin(282^{\circ} - 4x) = sin(222^{\circ} - 2x) - cos(2x - 72^{\circ})$ Now defining $y = 2x - 72^{\circ}$, $cos(2x - 72^{\circ}) = cosy$ $sin(222^{\circ} - 2x) = sin(150^{\circ} - y) = sin150^{\circ}cosy - cos150^{\circ}siny = \frac{1}{2}cosy + \frac{\sqrt3}{2}siny$ So, $sin(282^{\circ} - 4x) = \frac{\sqrt3}{2}siny - \frac{1}{2}cosy = sinycos30^{\circ} - cosysin30^{\circ} = sin(y - 30^{\circ}) = sin(2x - 102^{\circ})$ $\implies 282^{\circ} - 4x = n\pi \pm (2x - 102^{\circ})$ Trivial inequality calculations show that $n = 0$ is the only possible case, resulting in $x = 64^{\circ}$ Then finally, $\angle BAP = \boxed{13^{\circ}}$

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