Solution:

Answer: $8+4 \sqrt{2}$

We first maximize the area of triangle $P_{1} O P_{2}$, noting that the sum of the area of $P_{1} O P_{2}$ and the three other analogous triangles is the area of $P_{1} P_{2} P_{3} P_{4}$. Note that if $A \neq P_{1}, P_{2}$, without loss of generality say $\angle O A P_{1}<90^{\circ}$. Then, $\angle O O_{1} P_{1}=2 \angle O A P_{1}$, and since $\angle O A P_{2}=180^{\circ}-\angle O A P_{1}>90^{\circ}$, we see that $\angle O O_{2} P_{2}=2 \angle O A P_{1}$ as well, and it follows that $O O_{1} P_{1} \sim O O_{2} P_{2}$. This is a spiral similarity, so $O O_{1} O_{2} \sim O P_{1} P_{2}$, and in particular $\angle P_{1} O P_{2}=\angle O_{1} O O_{2}$, which is fixed. By the sine area formula, to maximize $O P_{1} \cdot O P_{2}$, which is bounded above by the diameters $2(O O_{1})$, $2(O O_{2})$. In a similar way, we want $P_{3}, P_{4}$ to be diametrically opposite $O_{3}, O_{4}$ in their respective circles.

When we take these $P_{i}$, we indeed have $A \in P_{1} P_{2}$ and similarly for $B, C, D$, since $\angle O A P_{1}=\angle O A P_{2}=90^{\circ}$. To finish, the area of the quadrilateral is the sum of the areas of the four triangles, which is

$$\frac{1}{2} \cdot \frac{\sqrt{2}}{2} \cdot 2^{2} \cdot (1 \cdot 2 + 2 \cdot \sqrt{2} + \sqrt{2} \cdot 2 + 2 \cdot 1) = 8 + 4 \sqrt{2}.$$