The projection of a parallelogram is always another parallelogram (or, in the flattest case, a line segment). It can never be a quadrilateral that is not a parallelogram.
Start where the hint suggests, with a pair of parallel lines. Two parallel lines and the direction of projection lie in two parallel planes. Those planes meet the projection plane in two parallel lines. So parallel lines project to parallel lines.
AB ∥ DC ⇒ their projections A′B′ ∥ D′C′
AD ∥ BC ⇒ their projections A′D′ ∥ B′C′
Both pairs of opposite sides parallel ⇒ A′B′C′D′ is a parallelogram
So a trapezium, a kite or an ordinary quadrilateral can never be the shadow of a parallelogram.
Try This: Cut out a parallelogram and look at its shadow in sunlight, which is as good as a projection because the Sun’s rays are effectively parallel. However you turn it, the shadow stays a parallelogram — it may become a rectangle, a square, a rhombus or a very thin sliver, but the opposite sides never stop being parallel.