Because the area of $ABDC$ is constant, finding the largest possible are of $ABDPC$ is equivalent to finding the minimal area of $\triangle{CDP}$. Let $\ell$ be a line of tangent of circle $O$ which passes point $P$. It is easy to see when $\ell\parallel CD$, the area of $\triangle{CDP}$ reaches minimal. When $\ell\parallel CD$, it must hold that $OP \perp CD$ because $OP \perp \ell$. It follows that the distance between $P$ and $CD$ must equal the difference between the distance from $O$ to $CD$ and circle $O$'s radius. Let's establish a coordinate system as shown where $O$ is the origin, $AB$ is the $x$-axis. Then the coordinates of $C$ and $D$ will be $(-1, 1)$ and $(1, 2)$, respectively. Therefore the equation of $CD$ is $$\frac{y-1}{x+1}=\frac{2-1}{1+1}\implies x-2y+3=0$$

The distance from $O$ to $CD$ equals $$\frac{|0-2\times 0 + 3|}{\sqrt{1^2+(-2)^2}}=\frac{3}{\sqrt{5}}$$ The length of $CD$ equal $\sqrt{(1+1)^2 + (2-1)^2}=\sqrt{5}$, hence $$S_{\triangle{CDP}}=\frac{1}{2}\times\sqrt{5}\times(\frac{3}{\sqrt{5}}-1)=\frac{3-\sqrt{5}}{2}$$ Meanwhile, $$S_{ABDC}=\frac{1}{2}\cdot (AC+BD)\cdot AB = \frac{1}{2}\times(1+2)\times 2=3$$ Therefore, the largest possible area of $ABDPC$ must be $$3-\frac{3-\sqrt{5}}{2}=\boxed{\frac{3+\sqrt{5}}{2}}$$