A circle with center at the origin and radius $5$ intersects the abscissa in points $A$ and $B$. Let $P$  a point lying on the line $x = 11$, and the point $Q$ is the intersection point of $AP$ with this circle. We know what is the $Q$ point is the midpoint of the $AP$. Find the coordinates of the point $P$.
1. The circle with center at the origin and radius 5 has the equation:
   \[
   x^2 + y^2 = 25
   \]
   The points where this circle intersects the x-axis (abscissa) are \(A(-5, 0)\) and \(B(5, 0)\).

2. Let \(P\) be a point on the line \(x = 11\). Therefore, the coordinates of \(P\) are \((11, \lambda)\).

3. The line \(AP\) intersects the circle at another point \(Q\). Given that \(Q\) is the midpoint of \(AP\), we need to find the coordinates of \(P\).

4. The coordinates of \(Q\) can be expressed as:
   \[
   Q\left(\frac{5(256 - \lambda^2)}{256 + \lambda^2}, y_Q\right)
   \]
   where \(y_Q\) is the y-coordinate of \(Q\).

5. Since \(Q\) is the midpoint of \(AP\), the midpoint formula gives:
   \[
   \left(\frac{-5 + 11}{2}, \frac{0 + \lambda}{2}\right) = \left(3, \frac{\lambda}{2}\right)
   \]

6. Equating the x-coordinates of the midpoint:
   \[
   \frac{5(256 - \lambda^2)}{256 + \lambda^2} = 3
   \]

7. Solving the equation:
   \[
   5(256 - \lambda^2) = 3(256 + \lambda^2)
   \]
   \[
   1280 - 5\lambda^2 = 768 + 3\lambda^2
   \]
   \[
   1280 - 768 = 8\lambda^2
   \]
   \[
   512 = 8\lambda^2
   \]
   \[
   \lambda^2 = 64
   \]
   \[
   \lambda = \pm 8
   \]

8. Therefore, the coordinates of \(P\) are \((11, 8)\) or \((11, -8)\).

The final answer is \( \boxed{ (11, 8) } \) or \((11, -8)\).