Both lines $AB$ and $DC$ must be parallel to p, since if one of them is not, then neither can be and they must both intersect $p$ (since they are both coplanar with $p$), making them skew.

Now we note since a circle can be inscribed in the trapezoid/trapezium, we must have $AB + DC = AD + BC$, and since the trapezoid/trapezium is isosceles, this implies that each of the trapezoid’s/trapezium’s legs has length equal to the average of the lengths of the bases.

We can find this average by dropping perpendicular $AA’$ to $DC$ such that $A’$ is on $DC$. The average will be $A’C$, which is one of the sides of the rectangle with sides on $AB$ and $DC$ with vertices at $A$ and $C$.

We now draw a circle with center $C$ that contains $A’$. The intersections of this circle with $AB$ are the two possible values of $B$, from either of which it is trivial to determine the corresponding location for $D$. It is worth noting that the intersection points may concur (in which case there is only one distinct possibility, a square), or they may not occur at all.