Two planes, $P$ and $Q$, intersect along the line $p$. The point $A$ is given in the plane $P$, and the point $C$ in the plane $Q$; neither of these points lies on the straight line $p$. Construct an isosceles trapezoid $ABCD$ (with $AB \parallel CD$) in which a circle can be inscribed, and with vertices $B$ and $D$ lying in planes $P$ and $Q$ respectively.
1. **Identify the given elements and their relationships:**
   - Two planes \( P \) and \( Q \) intersect along the line \( p \).
   - Point \( A \) is in plane \( P \) and point \( C \) is in plane \( Q \).
   - Neither \( A \) nor \( C \) lies on the line \( p \).
   - We need to construct an isosceles trapezoid \( ABCD \) with \( AB \parallel CD \) and vertices \( B \) and \( D \) lying in planes \( P \) and \( Q \) respectively.
   - The trapezoid must have an inscribed circle.

2. **Establish the properties of the trapezoid:**
   - Since \( AB \parallel CD \), both \( AB \) and \( CD \) must be parallel to the line \( p \).
   - The distance \( CE \) between the parallel sides \( AB \) and \( CD \) is equal to the diameter of the inscribed circle.
   - The distance \( AC \) between points \( A \) and \( C \) is known.
   - Denote the lengths of \( AB \) and \( CD \) by \( a \) and \( b \) respectively.
   - Denote the equal lengths of \( AD \) and \( BC \) by \( s \).

3. **Use the property of tangents from a point to a circle:**
   - For a quadrilateral to have an inscribed circle, the sum of the lengths of opposite sides must be equal.
   - Therefore, \( AB + CD = AD + BC \), which implies \( a + b = 2s \).
   - Hence, \( s = \frac{1}{2}(a + b) \).

4. **Determine the distances \( AE \) and \( BE \):**
   - If \( a < b \), then \( AE = a + BE = \frac{1}{2}(a + b) = s \).
   - If \( a > b \), then \( a - 2BE = b \), so \( BE = \frac{1}{2}(a - b) \) and \( AE = a - BE = \frac{1}{2}(a + b) = s \).
   - Thus, in both cases, \( AE = s \).

5. **Construct the trapezoid:**
   - In plane \( P \), construct a line \( l \) through \( A \) parallel to \( p \).
   - In plane \( Q \), construct a line \( m \) through \( C \) parallel to \( p \).
   - Construct a perpendicular to \( CD \) at \( C \) meeting \( l \) at \( E \).
   - With \( AE \) as the radius and \( C \) as the center, draw an arc of a circle intersecting line \( AE \) at \( B \).
   - With the same radius and \( A \) as the center, draw an arc of a circle intersecting line \( m \) at \( D \).
   - The points \( B \) and \( D \) are the desired vertices.

6. **Consider the conditions for the solution:**
   - If \( CE < AE \), two solutions exist.
   - If \( AE = CE \), there is only one solution, which is a square.
   - If \( CE > AE \), the problem has no solution.

The final answer is the construction of the isosceles trapezoid \( \boxed{ ABCD } \) with the given properties.