Thank you to Mayank, Campion School, Bhopal, India; Yung, from Hong Kong and Ruth from Manchester High School for Girls for your solutions. Here is Ruth's solution:

My conjecture is that PQRS is always a parallelogram. It seems
that PQ and RS are always perpendicular to BD and that QR and SP
are perpendicular to AC. If this is the case then PQRS is a
paralelogram because PQ and RS are perpendicular to the same line
so must be parallel. Similarly QR and PS are parallel so PQRS has
opposite sides parallel so is a parallelogram.

So to show PQRS is a parallelogram it suffices to show that
the line joining the centres of two circles (a line such as PQ) is
perpendicular to the common chord of these two circles (a line such
as BX.

Consider the triangles BQP and XQP. Sides BQ and XQ are both
radii of the circle centre Q so are equal length. Sides BP and XP
are both radii of the circle centre P so are equal length. Side QP
is common to both triangles. Therefore BQP and XQP are congruent by
SSS. Therefore quadrilateral BPXQ is symetrical along PQ. Therefore
BX is perpendicular to PQ as required. Therefore PQRS is a
parallelogram.

Chi Kin, St Dominic's International School of Lisbon, also gave an excellent proof including a discussion of the degenerate cases where one vertex, A, B, C or D is moved on top of another or on top of the point X. In these cases there is no longer a convex quadrilateral ABCD.

For instance, if we move C on top of B, both the points B, C and X are joined at one point. As a result, the circumcircle of the triangle BXC is reduced to a point, and only 3 circles will be left. Quadrilateral PQRS cannot be formed any more.

Consider also, as we move C on top of X, the common chord CX will be eventually reduced to a point, and RQ will therefore disappear. Quadrilateral PQRS can't be formed.

Finally, if we move C on top of A, two of the circles will be overlapping the other two. As a result, RQ will also overlap PS, and the quadrilateral PQRS is reduced to a line.