Find the side length of the equilateral triangle

Begin by rotating the hole figure by $60^\circ$ clockwise, about $C$ , as shown below.

$CP=C{P}'=2$ and $\angle PC{P}'=60{}^\circ$ , hence $\triangle PC{P}'$ is equilateral.
Thus, $P{P}'=2$ . Consequently, $\triangle AP{P}'$ is a $1$ - $\sqrt{3}$ - $2$ triangle, i.e. a $30{}^\circ$ - $60{}^\circ$ - $90{}^\circ$ right triangle, with $\angle A{P}'P=30{}^\circ$ .

Now, $\angle A{P}'C=\angle A{P}'P+\angle P{P}'C=30{}^\circ +60{}^\circ =90{}^\circ$ By Pytghagorean theorem on $\triangle A{P}'C$ , $A{{C}^{2}}=A{{{P}'}^{2}}+{P}'{{C}^{2}}\Rightarrow {{s}^{2}}={{\sqrt{3}}^{2}}+{{2}^{2}}=7\Rightarrow s=\boxed{\sqrt{7}}$

Begin by rotating $\triangle ABC$ by $60^\circ$ about $A$ , such that in $\triangle A'B'C'$ , $B'=C$ . We see that $triangle APP'$ is equilateral triangle with side length $1$ , meaning that $\angle APP'=60^\circ$ . We also see that $\triangle CPP'$ is a $30^\circ-60^\circ-90^\circ$ right triangle, meaning that $\angle CPP'=60^\circ$ . Thus, by adding the two together, we see that $\angle APC=120^\circ$ . We can now use the law of cosines as following:

$s^2 = (AP)^2 + (CP)^2 - 2(AP)(CP)(\cos \angle APC)$

$s^2=1+4-2(1)(2)\left(\frac{-1}{2}\right)$

$s^2=5-4\left(\frac{-1}{2}\right)$

$\large{\color{#3D99F6}\boxed{s=\sqrt{7}}}$ $\color{#EC7300}\text{answer}$

Let the centroid of $\triangle ABC$ be $O(0,0)$ , the origin of the $xy$ -plane. Then $A(0, \frac s {\sqrt 3} )$ , $B(-\frac s2, - \frac s{2 \sqrt 3})$ , and $C(\frac s2, - \frac s{2 \sqrt 3})$ . Let $P(x, y)$ . Then we have

$\begin {cases} x^2 +(y-\frac s{\sqrt 3})^2 = 1 &... (1) \\ (x+\frac s 2)^2 + (y+\frac s{2 \sqrt 3})^2 = 3 &... (2) \\ (x-\frac s 2)^2 + (y+\frac s{2 \sqrt 3})^2 = 4 &... (3) \end {cases}$

From $(3)-(2): \ - 2sx=1 \implies x =-\frac 1{2 s}$ .

From $(2)-(1)$ :

$\begin{aligned} sx +\frac {s^2} 4+\frac {sy} {\sqrt 3}+\frac {s^2} {1 2}+ \frac {2sy}{\sqrt 3}-\frac {s^2} 3&=2 \\ - \frac 12 +\sqrt 3sy & =2 \\ \implies y&= \frac 5{2 \sqrt 3s} \end{aligned}$

Substitute $x$ and $y$ in $(1)$ :

$\begin{aligned} \frac 1{4s^2} +\left (\frac 5{2 \sqrt 3s}-\frac s{\sqrt 3} \right) ^2 & =1 \\ \frac 1{4s^2} +\frac {2 5}{1 2s^2}-\frac 53+\frac {s^2}3 & =1 \\ \frac 7{3s^2} - \frac 83+\frac {s^2} 3&=0 \\ s^4 - 8s^2 +7&=0 \\ (s^2 - 1)(s^2 - 7)&=0 & \small \blue {\text {Since } s >1} \\ \implies s & = \boxed{ \sqrt 7} \end{aligned}$

Let $\angle {PCA}=α,\angle {PCB}=60\degree-α$

Then

$\cos α=\dfrac {s^2+3}{4s}$

$\cos (60\degree-α)=\dfrac {s^2+1}{4s}$

Solving we get

$\sin α=\dfrac {s^2-1}{4\sqrt 3s}$

So, $\cos^2 α+\sin^2 α=1\implies \dfrac {(s^2+3)^2}{16s^2}+\dfrac {(s^2-1)^2}{48s^2}=1$

$\implies s^4-8s^2+7=0\implies s=1,\sqrt 7$

Obviously $s$ can't be $1$ , since then $|\overline {PA}|+|\overline {AC}|=1+1=2=|\overline {PC}|$

Therefore $s=\boxed {\sqrt 7}$ .