My favorite interpretation comes from the following claim / 'obvious fact'
Claim: Amongst all smooth curves with a fixed perimeter $P$, the area $A$ satisfies the inequality $ 4 \pi A \leq P^2$. Equality holds when the curve is a circle.
Lemma: If the 4 given sides satisfy the cyclic inequalities $a < b+c+d$, then a quadrilateral with sides $a, b, c, d$ exists. Moreover, a cyclic quadrilateral with sides $a, b, c, d$ exists. A sketch of this is given at the end.
This Lemma merely serves to let us know when the question makes geometric sense. Now, let $ABCD$ be a cyclic quad with the desired side lengths, and consider it's circumcircle. Let it have area $X$. Fix the circular arcs along with the intermediate areas between the circular arcs and the quadrilateral. Let these arcs have perimeter $P$, and the intermediate area have total area $I$.
Take any deformation $A^*B^*C^*D^*$, where the arcs and the areas are moved under translation and rotation. This gives a figure with perimeter $P$, and area $I+X^*$, where $X^*$ is the area of the deformed $A^*B^*C^*D^*$. By the initial claim, $I+X^* \leq I+X$. Hence we are done.$ _ \square$
Sketch Proof of Lemma: The first part follows from the triangle inequality. The second part follows by taking a continuous deformation of the quadrilateral, showing that opposite angles must equal $180^\circ$ at some point in time.
You found $DQ=2$ correctly. Now you must find $BP$. Apply Cosine theorem for $\triangle ABD$:
$$AD^2=AB^2+BD^2-2AB\cdot BD\cdot \cos ABD \Rightarrow \\
\cos ABD=\frac{4^2+8^2-6^2}{2\cdot 4\cdot 8}=\frac{11}{16} \Rightarrow \\
\cos (2\cdot \frac{ABD}2)=2\cos^2(\frac{ABD}2)-1=\frac{11}{16} \Rightarrow \\
\cos^2(\frac{ABD}2)=\frac{27}{32}=\frac{1}{1+\tan^2 (\frac{ABD}2)}\Rightarrow \\
\tan (\frac{ABD}2)=\frac{\sqrt{5}}{3\sqrt{3}}=\frac{C_1P}{BP}\Rightarrow \\
BP=\frac{3\sqrt{3}\cdot \frac{\sqrt{15}}{3}}{\sqrt{5}}=3$$
Hence, $PQ=BD-BP-DQ=8-3-2=3$.
Can you finish it yourself?
Answer:
$A=\frac12\cdot PQ\cdot\left(C_1P+C_2Q\right)=\frac12\cdot 3\cdot \left(\frac{\sqrt{15}}{3}+2\right).$
Best Answer
Consider $\triangle PAB$ and $\triangle PDC$. These pairs of inscribed angles are equal:
$$\begin{align*} \angle CAB &=\angle BDC\\ \angle ABD &= \angle DCA \end{align*}$$
So $\triangle PAB \sim\triangle PDC$. The ratio of their areas are square of the ratio of their corresponding sides:
$$\frac{S_{\triangle PAB}}{S_{\triangle PDC}} = \left(\frac{AB}{DC}\right)^2 = \frac{1}{81}$$
Similar, for $\triangle PBC$ and $\triangle PAD$, consider these pairs of inscribed angles,
$$\begin{align*} \angle DBC &= \angle CAD\\ \angle BCA &= \angle ADB\\ \triangle PBC &\sim \triangle PAD\\ \frac{S_{\triangle PBC}}{S_{\triangle PAD}} &= \left(\frac{BC}{AD}\right)^2\\ & = \frac{81}{64} \end{align*}$$
Between $\triangle PAB$ and $\triangle PAD$, these inscribed angles are equal because $BC = CD$:
$$\angle CAB = \angle CAD$$
Then the ratio of their areas is
$$\begin{align*} \frac{S_{\triangle PAB}}{S_{\triangle PAD}} &= \frac{\frac12 PA\cdot AB\sin\angle PAB}{\frac12PA\cdot AD\sin PAD}\\ &= \frac{AB}{AD}\\ &= \frac 18\\ \end{align*}$$
$$S_{\triangle PAB}:S_{\triangle PBC}:S_{\triangle PCD}:S_{\triangle PDA} = 8:81:648:64 $$