An isosceles trapezoid with bases $a$ and $c$ and altitude $h$ is given.

a) On the axis of symmetry of this trapezoid, find all points $P$ such that both legs of the trapezoid subtend right angles at $P$;

b) Calculate the distance of $p$ from either base;

c) Determine under what conditions such points $P$ actually exist. Discuss various cases that might arise.
Given an isosceles trapezoid with bases \( a \) and \( c \) and altitude \( h \), we need to find points \( P \) on the axis of symmetry such that both legs of the trapezoid subtend right angles at \( P \).

### Part (a)

1. **Identify the Trapezoid and Symmetry Axis:**
   Let the trapezoid be \( ABCD \) with \( AB = a \), \( CD = c \), and \( AD = BC \). The axis of symmetry is the vertical line passing through the midpoints of \( AB \) and \( CD \).

2. **Construct the Circle:**
   Consider the circle with diameter \( AD \). Since \( AD = BC \), this circle will also have diameter \( BC \). The points where this circle intersects the axis of symmetry are the points \( P \) we are looking for.

3. **Right Angles at \( P \):**
   By the Inscribed Angle Theorem, any angle subtended by a diameter of a circle is a right angle. Therefore, the points \( P \) where the circle intersects the axis of symmetry will subtend right angles at both \( AD \) and \( BC \).

### Part (b)

1. **Distance Calculation:**
   Let \( M \) and \( N \) be the midpoints of \( AB \) and \( CD \), respectively. The axis of symmetry is the line \( MN \).

2. **Coordinates of Points:**
   Place the trapezoid in the coordinate plane with \( M \) at \( (0, 0) \) and \( N \) at \( (0, h) \). The points \( P \) will be on the line \( MN \).

3. **Distance from \( P \) to Bases:**
   Let \( P \) be at \( (0, y) \). The distance from \( P \) to the base \( AB \) is \( |y| \) and the distance from \( P \) to the base \( CD \) is \( |h - y| \).

### Part (c)

1. **Existence of Points \( P \):**
   For points \( P \) to exist, the circle with diameter \( AD \) must intersect the axis of symmetry. The radius of this circle is \( \frac{AD}{2} \).

2. **Condition for Intersection:**
   The distance from the center of the circle (midpoint of \( AD \)) to the axis of symmetry must be less than or equal to the radius. This gives the condition:
   \[
   \sqrt{\left(\frac{a - c}{2}\right)^2 + h^2} \leq \frac{a + c}{2}
   \]

3. **Simplify the Condition:**
   \[
   \left(\frac{a - c}{2}\right)^2 + h^2 \leq \left(\frac{a + c}{2}\right)^2
   \]
   \[
   \frac{(a - c)^2}{4} + h^2 \leq \frac{(a + c)^2}{4}
   \]
   \[
   (a - c)^2 + 4h^2 \leq (a + c)^2
   \]
   \[
   a^2 - 2ac + c^2 + 4h^2 \leq a^2 + 2ac + c^2
   \]
   \[
   4h^2 \leq 4ac
   \]
   \[
   h^2 \leq ac
   \]

Thus, the points \( P \) exist if and only if \( h^2 \leq ac \).

\(\blacksquare\)

The final answer is \( \boxed{ h^2 \leq ac } \)