There are several possible configurations depending on the location of $P$ and the orientation of $\gamma$. We will consider the configuration above but will use directed lengths and angles so our arguments apply to all diagram configurations. By the law of sines on triangles $A_1PB$ and $A_1CP$, we have
$$BP \cdot \frac{\sin \angle A_1 PB}{BA_1} = \sin \angle BA_1 P = \sin \angle PA_1 C = CP \cdot \frac{\sin \angle CPA_1}{CA_1}.$$
Rearranging and considering the two other analogous equalities yields
$$\frac{BA_1}{A_1C} = -\frac{BP \sin \angle A_1 PB}{CP \sin \angle CPA_1}, \quad \frac{CB_1}{B_1A} = -\frac{AP \sin \angle C_1 PA}{BP \sin \angle BPC_1}, \quad \text{and} \quad \frac{AC_1}{C_1B} = -\frac{CP \sin \angle B_1 PC}{AP \sin \angle APB_1}. \quad (9)$$

Now, observe that $\angle CPA_1$ and $\angle C_1PA$ are angles between lines which are reflections of each other, meaning that they are either equal or supplementary. In either case, applying analogous arguments, we obtain
$$\sin \angle C_1 PA = \sin \angle CPA_1, \quad \sin \angle A_1 PB = \sin \angle APB_1, \quad \text{and} \quad \sin \angle B_1 PC = \sin \angle BPC_1.$$

Put the points on the complex plane and denote the complex number representing a point by the corresponding lowercase letter. Place $P$ at the origin and let $\gamma$ be the real line, so that reflection about $\gamma$ is given by complex conjugation. Now, because $A_1$ lies on $BC$, we have
$$\frac{a_1 - b}{b - c} = \frac{\bar{a}_1 - \bar{b}}{\bar{b} - \bar{c}},$$
while the fact that it lies on the reflection of $AP$ implies that it is proportional to $\bar{a}$, meaning that $\frac{a_1}{\bar{a}} = \frac{\bar{a}_1}{\bar{a}}$. Substituting into the previous equation and solving for $a_1$ yields
$$a_1 = \frac{\bar{a}\bar{b}c - \bar{b}\bar{c}a}{\bar{a}(\bar{b}\bar{c} - \bar{c}) - a(b - c)}.$$
Now, because $A_1$ lies on $BC$, we may compute
$$\frac{BA_1}{A_1C} = \frac{b - a_1}{a_1 - c} = \frac{\bar{a}b(\bar{b} - \bar{c}) - ab(b - c) - \bar{a}(\bar{b}c - \bar{c}b)}{\bar{a}(\bar{b}c - \bar{c}b) - \bar{a}c(\bar{b} - \bar{c}) + ac(b - c)} = \frac{(\bar{a}\bar{b} - ab)(b - c)}{(\bar{a}c - ac)(c - b)}.$$
Multiplying this with the analogous expressions for $\frac{CB_1}{B_1A}$ and $\frac{AC_1}{C_1B}$, we obtain
$$\frac{BA_1}{A_1C} \cdot \frac{CB_1}{B_1A} \cdot \frac{AC_1}{C_1B} = -1,$$
again yielding the conclusion by Menelaus' theorem.