IMO Shortlist, 2006

Because $AB \parallel CD$, the relation $AK / KB = DL / LC$ readily implies that the lines $AD$, $BC$ and $KL$ have a common point $S$.

Consider the second intersection points $X$ and $Y$ of the line $SK$ with the circles $(ABP)$ and $(CDQ)$, respectively. Since $APBX$ is a cyclic quadrilateral and $AB \parallel CD$, one has
$$\angle AXB = 180^{\circ} - \angle APB = 180^{\circ} - \angle BCD = \angle ABC.$$
This shows that $BC$ is tangent to the circle $(ABP)$ at $B$. Likewise, $BC$ is tangent to the circle $(CDQ)$ at $C$. Therefore $SP \cdot SX = SB^{2}$ and $SQ \cdot SY = SC^{2}$.
Let $h$ be the homothety with centre $S$ and ratio $SC / SB$. Since $h(B) = C$, the above conclusion about tangency implies that $h$ takes circle $(ABP)$ to circle $(CDQ)$. Also, $h$ takes $AB$ to $CD$, and it easily follows that $h(P) = Y$, $h(X) = Q$, yielding $SP / SY = SB / SC = SX / SQ$.
Equalities $SP \cdot SX = SB^{2}$ and $SQ / SX = SC / SB$ imply $SP \cdot SQ = SB \cdot SC$, which is equivalent to $P$, $Q$, $B$ and $C$ being concyclic.

The case where $P = Q$ is trivial. Thus assume that $P$ and $Q$ are two distinct points. As in the first solution, notice that the lines $AD$, $BC$ and $KL$ concur at a point $S$.

Let the lines $AP$ and $DQ$ meet at $E$, and let $BP$ and $CQ$ meet at $F$. Then $\angle EPF = \angle BCD$ and $\angle FQE = \angle ABC$ by the condition of the problem. Since the angles $BCD$ and $ABC$ add up to $180^{\circ}$, it follows that $PEQF$ is a cyclic quadrilateral.
Applying Menelaus' theorem, first to triangle $ASP$ and line $DQ$ and then to triangle $BSP$ and line $CQ$, we have
$$\frac{AD}{DS} \cdot \frac{SQ}{QP} \cdot \frac{PE}{EA} = 1 \quad \text{and} \quad \frac{BC}{CS} \cdot \frac{SQ}{QP} \cdot \frac{PF}{FB} = 1.$$
The first factors in these equations are equal, as $AB \parallel CD$. Thus the last factors are also equal, which implies that $EF$ is parallel to $AB$ and $CD$. Using this and the cyclicity of $PEQF$, we obtain
$$\angle BCD = \angle BCF + \angle FCD = \angle BCQ + \angle EFQ = \angle BCQ + \angle EPQ.$$
On the other hand,
$$\angle BCD = \angle APB = \angle EPF = \angle EPQ + \angle QPF,$$
and consequently $\angle BCQ = \angle QPF$. The latter angle either coincides with $\angle QPB$ or is supplementary to $\angle QPB$, depending on whether $Q$ lies between $K$ and $P$ or not. In either case it follows that $P$, $Q$, $B$ and $C$ are concyclic.