Circle $\Gamma$ with radius $1$ is centered at point $A$ on the circumference of circle $\omega$ with radius $7$. Suppose that point $P$ lies on $\omega$ with $AP=4$. Determine the product of the distances from $P$ to the two intersections of $\omega$ and $\Gamma$.

[i]2018 CCA Math Bonanza Team Round #6[/i]
1. Let $\Gamma$ and $\omega$ intersect at points $X$ and $Y$. We need to determine the product of the distances from point $P$ to these intersection points.

2. Since $A$ is the center of $\Gamma$ and lies on the circumference of $\omega$, the distance $AP$ is given as $4$. The radius of $\omega$ is $7$, and the radius of $\Gamma$ is $1$.

3. By the Power of a Point theorem, for a point $P$ outside a circle, the power of the point is given by the product of the distances from $P$ to the points of intersection of any line through $P$ with the circle. In this case, we apply the Power of a Point theorem to point $P$ with respect to circle $\Gamma$.

4. The power of point $P$ with respect to circle $\Gamma$ is given by:
   \[
   PA^2 - r^2
   \]
   where $PA$ is the distance from $P$ to the center of $\Gamma$ (which is $A$), and $r$ is the radius of $\Gamma$.

5. Substituting the given values, we have:
   \[
   PA = 4 \quad \text{and} \quad r = 1
   \]
   Therefore, the power of point $P$ with respect to circle $\Gamma$ is:
   \[
   PA^2 - r^2 = 4^2 - 1^2 = 16 - 1 = 15
   \]

6. By the Power of a Point theorem, the product of the distances from $P$ to the points of intersection $X$ and $Y$ is equal to the power of point $P$ with respect to circle $\Gamma$. Hence:
   \[
   PX \cdot PY = 15
   \]

Conclusion:
\[
\boxed{15}
\]