Let ABC be a triangle with P a point on line BC
such that B lies between P and C.

Construct a line through P, intersecting sides
AB and AC in points Q and R respectively, such 
that vertex S of parallelogram QARS lies on 
line BC.

The problem asks to find a point P on line BC such that B is between P and C. If I remember some basic geometry correctly, it's impossible to have a point on the end of a line be between two other points on the line, because then it would no longer be an end.

The closest we can get to that is if P is at B. Then the only way there can be a line through P that intersects points on AB and AC is if the angle at A is 180 degrees. This makes the angles at B and C 0 degrees. This is a special case of a triangle where the triangle is a line. If all other vertexes of QARS are on this line, then S is also on this line, hence, vertex on a line.