A Cyclic Quadrilateral

Let $PC = x$ and $\angle APC = \theta$ . Then, by sine rule:

$\begin{cases} \dfrac{x}{\sin 90^\circ} = \dfrac{2}{\sin \theta} & \Rightarrow x = \dfrac{2}{\sin \theta} & ...(1) \\ \dfrac{x}{\sin 90^\circ} = \dfrac{3}{\sin (60^\circ-\theta)} & \Rightarrow x = \dfrac{3}{\sin (60^\circ-\theta)} & ...(2) \end{cases}$

$\begin{aligned} \frac{3}{\sin (60^\circ-\theta)} & = \frac{2}{\sin \theta} \\ 3 \sin \theta & = 2\sin (60^\circ-\theta) \\ 3 \sin \theta & = 2 \left(\frac{\sqrt{3}}{2} \cos \theta - \frac{1}{2} \sin \theta \right) \\ 3 \sin \theta & = \sqrt{3} \cos \theta - \sin \theta \\ 4 \sin \theta & = \sqrt{3}\cos \theta \\ \Rightarrow \tan \theta & = \frac{\sqrt{3}}{4} \\ \Rightarrow \sin \theta & = \sqrt{\frac{3}{19}} \\ \\ (1): \quad x & = \frac{2}{\sin \theta} = 2\sqrt{\frac{19}{3}} \approx \boxed{5.033} \end{aligned}$

$AB = x \Rightarrow x^{2} = AC^{2} +CB^{2} - 2\cdot AB\cdot BC\cdot \cos (120º) = 4+9+6=19$ $AP = y, PB = z \Rightarrow z^{2}+y^{2}-2yz \cos (60º) =x^{2} \Rightarrow z^{2}+y^{2}-yz = 19$
$PC^{2} = z^{2}+3^{2} = y^{2}+2^{2} \Rightarrow y^{2}=5+z^{2}$ $\Rightarrow z^{2}+y^{2}-yz = z^{2}+(5+z^{2})-z(\sqrt{5+z^{2}}) =19$ $\Rightarrow z = \frac{7}{\sqrt{3}}$ $\Rightarrow PC = \sqrt{z^{2}+3^{2}}=\sqrt{\frac{49}{3}+9}=\frac{2\sqrt{57}}{3}\approx 5.033$