We will use complex numbers and suppose that $\Gamma$ is centred at the origin so that $\Gamma$ is the set of complex numbers $w$ such that $|w| = 1$. Moreover, w.l.o.g., we can suppose that $A = 1$, $B = \omega$, $C = \omega^2$, where $\omega = e^{2\pi i/3}$ is a cube root of unity and $1 + \omega + \omega^2 = 0$. Let the complex number $z$ stand for $P$. Then
$$|PA| = |z - 1|, \quad |PB| = |z - \omega|, \quad |PC| = |z - \omega^2|,$$
and so
$$\begin{aligned} |PA| \cdot |PB| \cdot |PC| &= |z-1||z-\omega||z-\omega^2| \\ &= |(z-1)(z-\omega)(z-\omega^2)| \\ &= |z^3 - 1| \\ &\le |z^3| + 1 \quad (\text{since } |a+b| \le |a| + |b|) \\ &= |z|^3 + 1 \\ &\le 2 \end{aligned}$$
if $P$ is within or on $\Gamma$, in which case $|z| \le 1$.

We will now show that equality occurs exactly when $P$ is one of the three points on $\Gamma$ that are obtained by a $60^\circ$-rotation of $A, B, C$. Equality occurs above when $|z^3| = 1$ and $|z^3 - 1| = 2$, i.e. when $z^3$ is on $\Gamma$ and has distance $2$ from the complex number $1$. This happens exactly when $z^3 = -1$. This equation has three solutions: $-1, -\omega, -\omega^2$. These three complex numbers are diametrically opposite $A, B, C$ on $\Gamma$ and so can also be obtained by rotating $A, B, C$ by $60^\circ$ about the circumcentre of triangle $ABC$.