Option A: 4
To solve this problem, we need to determine the distance of the fly from the ceiling given the distances from the walls and the point P.
     Let's denote:
         ○ The distance from the fly to the first wall as $x = 1$ meter.
         ○ The distance from the fly to the second wall as $y = 8$ meters.
         ○ The distance from the fly to the point P as $z = 9$ meters.
         ○ The distance from the fly to the ceiling as $h$.
     Since the walls and the ceiling meet at right angles at point P, we can use the Pythagorean theorem in three dimensions. The fly's position can be represented as a point $(x, y, h)$ in a 3D coordinate system where point P is the origin $(0, 0, 0)$.
     The distance from the fly to point P is given by the formula for the distance between two points in 3D space:
     $$\sqrt{x^2 + y^2 + h^2} = z$$
     Substituting the known values:
     $$\sqrt{1^2 + 8^2 + h^2} = 9$$
     Simplifying inside the square root:
     $$\sqrt{1 + 64 + h^2} = 9$$
     $$\sqrt{65 + h^2} = 9$$
     Squaring both sides to eliminate the square root:
     $$65 + h^2 = 81$$
     Solving for $h^2$:
     $$h^2 = 81 - 65$$
     $$h^2 = 16$$
     Taking the square root of both sides:
     $$h = \sqrt{16} = 4$$
     Therefore, the fly is 4 meters from the ceiling.