Solution:

1.
The extensions of $A P, B P, C P$ meet the circumcircle of triangle $A B C$ again in $A'$, $B'$, $C'$. By means of the inscribed angle theorem and a simple angle-sum argument we find $\text{B' A' C' = B' A' A + A A' C' = B' B A + A C C' = B P C - B A C = 60}$. Likewise $\text{A' C' B' = 60}$ and hence the triangle $A' B' C'$ is equilateral, i.e. $A' B' = B' C' = C' A'$. As usual we have $A P B \sim B' P A'$ and $A P C \sim C' P A'$; from this it follows that $\frac{A P}{A C} = \frac{C' P}{C' A'}$ and $\frac{B P}{B C} = \frac{C' P}{C' B'}$. Combined with the previous, this teaches $\frac{A P}{A C} = \frac{B P}{B C}$ and thus indeed $A B \cdot P C = A C \cdot P B$.

2.
Choose the point $J$ such that the triangles $A B C$, $P B J$ (equally oriented) are similar. Then $\frac{A B}{B P} = \frac{B C}{B J}$ and $\text{P B A = J B C}$, wherefore the triangles $A B P$, $C B J$ (equally oriented) must also be similar. Consequently $\text{C J P = C J B - P J B = (60 + ) - = 60}$. Since also $\text{J P C = B P C - B P J = (60 + ) - = 60}$, the triangle $P C J$ is equilateral and hence $P C = P J$. By the choice of $J$ we have $\frac{A B}{A C} = \frac{P B}{P J}$ and from this it follows with the help of the previous, as desired, $A C \cdot P B = A B \cdot P J = A B \cdot P C$.

3.
Over the segment $B P$ the equilateral triangle $B P T$ is erected. The intersection point of $P T$ with $A B$ is called $G$. Furthermore, on $A C$ the point $H$ is chosen with $\text{C P H = 60}$. Because of $\text{A P G =}$ and $\text{H P A =}$ we must have $\text{H P G + G A H = 180}$, i.e. the quadrilateral $G A H P$ is inscribed in a circle. Accordingly $\text{H G A = H P A =}$, from which it is immediately concluded that $G H \parallel B C$. For this reason $\frac{A B}{A C} = \frac{B G}{C H}$ (1). Furthermore, simple angle considerations show $\text{H C P = 60 - P B G = G B T}$, which together with $\text{C P H = B T G [=60 ]}$ teaches the similarity of the triangles $P C H$ and $T B G$. Hence we have $\frac{B G}{C H} = \frac{B T}{C P}$ (2). Since the triangle $B P T$ is equilateral by construction, in particular $B T = B P$ holds. Together with (1) and (2) we obtain from this $\frac{A B}{A C} = \frac{B P}{C P}$ and thus indeed $A B \cdot P C = A C \cdot P B$.

4.
Choose the point $G$ such that the triangles $A G C$, $P B C$ (equally oriented) are similar. As in the second solution we see that then the triangles $G B C$ and $A P C$ (equally oriented) are also similar. Furthermore $\text{B A G = C A G - C A B = (60 + ) - = 60}$ and likewise $\text{G B A = 60}$. The triangle $A G B$ is thus equilateral and hence $A G = A B$. By $A G C \sim P B C$ we thus have $P B : P C = A G : A C = A B : A C$. Therefore, as claimed, $A B \cdot P C = A C \cdot P B$.

5.
Let the feet of the perpendiculars from $P$ onto the sides $B C$, $C A$, $A B$ be called $X, Y, Z$. By Thales' theorem the quadrilaterals $P X C Y$, $P Y A Z$, $P Z B X$ each have a circumcircle. We now find $\text{Z X Y = Z X P + P X Y = Z B P + P C Y = 60}$ and analogously $\text{X Y Z = Y Z X = 60}$. The triangle $X Y Z$ is thus equilateral. For the length of the side $X Y$ we find, by two applications of the law of sines, $X Y = P C \cdot \sin \gamma = \frac{A B \cdot P C}{2 R}$, where $R$ denotes the radius of the circumcircle of triangle $A B C$. Likewise $X Z = \frac{A C \cdot P B}{2 R}$. From $X Y = X Z$ it now follows, as required, $A B \cdot P C = A C \cdot P B$.

6.
The circumcircle of triangle $A B P$ intersects $A C$ a second time at $Q$. We obtain $\text{B Q A = B P A = + 60}$ and from this, by the exterior angle theorem, $\text{Q B C = 60}$. Moreover, from $\text{B P Q = 180 -}$ and $\text{C P B = 60 +}$ we immediately get $\text{Q P C = 120}$. Furthermore, if we set $\text{A B P =}$, we immediately obtain $\text{C Q P =}$. By repeated use of the law of sines we now obtain
$$\frac{P C}{P A} = \frac{P C}{C Q} \cdot \frac{C Q}{Q B} \cdot \frac{Q B}{A P} = \frac{\sin \varphi}{\sin 120^{\circ}} \cdot \frac{\sin 60^{\circ}}{\sin \gamma} \cdot \frac{\sin \alpha}{\sin \varphi} = \frac{\sin \alpha}{\sin \gamma} = \frac{B C}{A B}$$
From this it follows immediately $A B \cdot P C = A P \cdot B C$.

7.
Over the side $A B$ one erects the equilateral triangle $A B Q$ inward. Simple angle considerations now show $\text{Q A P = 60 - P A B = B C P}$ and likewise $\text{P B Q = P C A}$. If we now introduce into the consideration the intersection point of $A C$ with $B Q$ as $T$, the quadrilateral $B C T P$ will be a cyclic quadrilateral because of $\text{P B T = P B Q = P C A = P C T}$. Consequently $\text{B T P = B C P = Q A P}$, hence $\text{P T Q + Q A P = 180}$ and thus $P T Q A$ is also a cyclic quadrilateral. For this reason $\text{B Q P = T Q P = T A P = C A P}$ holds, from which, together with $\text{P B Q = P C A}$, the similarity of the triangles $B Q P$, $C A P$ follows. From this it is evident that $P C : A C = P B : B Q = P B : A B$, hence, as desired, $A B \cdot P C = A C \cdot P B$.

8.
Set $\text{A C P = '}$, $\text{P B C = ''}$, $\text{P B A = ''}$. Now by the law of sines
$$\frac{\sin \beta''}{\sin \gamma'} = \frac{\sin \beta''}{A P} \cdot \frac{A P}{\sin \gamma'} = \frac{\sin (60^{\circ} + \gamma)}{c} \cdot \frac{b}{\sin (60^{\circ} + \beta)} = \frac{\sin (60^{\circ} + \gamma)}{\sin \gamma} \cdot \frac{\sin \beta}{\sin (60^{\circ} + \beta)} = \frac{1 + \sqrt{3} \cot \gamma}{1 + \sqrt{3} \cot \beta}$$
Since however $\beta'' + \gamma' = 60^{\circ}$ we also have
$$\frac{\sin \beta''}{\sin \gamma'} = \frac{\sin (60^{\circ} - \gamma')}{\sin \gamma'} = \frac{\sqrt{3}}{2} \cot \gamma' - \frac{1}{2}$$
These two equations together yield
$$\cot \gamma' = \frac{\sqrt{3} + \cot \beta + 2 \cot \gamma}{1 + \sqrt{3} \cot \beta}$$
Furthermore
$$\frac{\sin \gamma''}{\sin \gamma'} = \frac{\sin (\gamma - \gamma')}{\sin \gamma'} = \sin \gamma \cot \gamma' - \cos \gamma$$
Herein we insert the equation found previously and obtain
$$\begin{gathered} \frac{\sin \gamma''}{\sin \gamma'} = \frac{(\sqrt{3} \sin \beta \sin \gamma + \cos \beta \sin \gamma + 2 \sin \beta \cos \gamma) - (\sin \beta \cos \gamma + \sqrt{3} \cos \beta \cos \gamma)}{\sin \beta + \sqrt{3} \cos \beta} \\ = \frac{\sin \alpha + \sqrt{3} \cos \alpha}{\sin \beta + \sqrt{3} \cos \beta} = \frac{\sin (60^{\circ} + \alpha)}{\sin (60^{\circ} + \beta)} \end{gathered}$$
Therefore by the law of sines
$$\frac{A P}{A C} = \frac{\sin \gamma'}{\sin (\beta + 60^{\circ})} = \frac{\sin \gamma''}{\sin (\alpha + 60^{\circ})} = \frac{B P}{B C}$$
i.e. as desired $A P \cdot B C = B P \cdot A C$.

9.
The circumcircle of triangle $A P B$ meets the line $C P$ a second time at $J$. Then $\text{A J P = A B P}$. Together with $\text{A B P + P C A = 60}$ this teaches $\text{C A J = 120}$. Likewise we see $\text{J B C = 120}$. Thus we have
$$\text{A C B C = A C J C J C B C = A J C C A J J B C C J B = A J P P J B = A B P P A B = A P B P ,}$$
i.e. as claimed $A C \cdot B P = B C \cdot A P$.