First solution. Let $E$ be the intersection point of diagonals $AC$ and $BD$. Assume for definiteness that point $K$ lies on segment $BE$. Let the line passing through $K$ and parallel to $PM$ intersect $AC$ at point $N$ (see Fig. 8). Then $\triangle NKE \sim \triangle PME$ (since their sides are parallel), hence $\frac{PE}{EM} = \frac{NE}{EK}$. On the other hand, the right triangles $AKE$ and $CME$ are also similar (by the acute angle at vertex $E$), so $\frac{EM}{EC} = \frac{EK}{EA}$. Multiplying the obtained equalities, we get $\frac{PE}{EC} = \frac{NE}{EA}$. But by Thales' theorem $\frac{PE}{EC} = \frac{KE}{EB}$.

Thus, we have $\frac{NE}{EA} = \frac{PE}{EC} = \frac{KE}{EB}$, whence $KN \parallel AB$. Therefore, $PM \parallel AB \perp BC \parallel KP$, as required.

Second solution. Again, let $E$ be the intersection point of diagonals $AC$ and $BD$ and consider the case when point $K$ lies on segment $BE$. Note that $\angle PAD = \angle CAD = \angle CBD = \angle PKD$, that is, quadrilateral $AKPD$ is cyclic (see Fig. 9). Therefore, $\angle AKD = \angle APD = 90^\circ$. Then from the relation $\angle CPD = \angle CMD = 90^\circ$ it follows that quadrilateral $CPMD$ is also cyclic, whence $\angle KPM = 180^\circ - \angle CPM = \angle EDC = \angle BAC = \angle BAE$. From this it follows that $PM \parallel AB \perp BC \parallel KP$, as required.

The case when $K$ lies on segment $DE$ is considered similarly.