The triangles $ACD$ and $ABC$ are similar, so $\angle CAD = \angle BAC$. We notice that the point $E$ does not lie on the same side of the line $BC$ as $A$, so $\angle BCE = 180^\circ - \angle ACB - \angle DCA = 90^\circ - \angle DCA = \angle CAD = \angle BAC$. We conclude that the triangles $ABC$ and $CEB$ are also similar since they have two congruent angles. By Pythagoras' theorem we have $|AB| = \sqrt{a^2 + b^2}$.

Similarity of the triangles $ABC$ and $ACD$ implies
$$\frac{|AD|}{|AC|} = \frac{|AC|}{|AB|} \quad \text{and} \quad \frac{|DC|}{|AC|} = \frac{|CB|}{|AB|}$$

From the first equality we get $|AD| = \frac{b^2}{\sqrt{a^2+b^2}}$, and from the second $|DC| = \frac{ab}{\sqrt{a^2+b^2}}$. Since the triangles $ABC$ and $CEB$ are also similar, we have
$$\frac{|BE|}{|BC|} = \frac{|CB|}{|CA|}$$
which implies $|BE| = \frac{a^2}{b}$.

Thus, the area of the quadrilateral $ABED$ equals
$$\begin{align*} p &= \frac{|AD| \cdot |DC|}{2} + \frac{|AC| \cdot |CB|}{2} + \frac{|CB| \cdot |BE|}{2} \\ &= \frac{ab^3}{2(a^2 + b^2)} + \frac{ab}{2} + \frac{a^3}{2b} = \frac{ab^4 + ab^2(a^2 + b^2) + a^3(a^2 + b^2)}{2b(a^2 + b^2)} \\ &= \frac{2ab^4 + 2a^3b^2 + a^5}{2b(a^2 + b^2)}. \end{align*}$$