Doubling the cube

$\begin{aligned} \text{As shown}&\text{ by the above figure let the paper be of side length } a\\ \text{Let } PA&= x\\ AC&=y\\\\ \implies PB&=a-x\\ CP&=a-y\\\\ \text{In } &\Delta ACP \text{ we have,}\\\\ x^2+y^2&=(a-y)^2\hspace{5mm}\color{#3D99F6}\small\angle PAC={90}^{\circ}\\ x^2&=a^2-2ay\\ y&=\dfrac{a^2-x^2}{2a}\\ a-y&=\dfrac{a^2+x^2}{2a}\\ \sin{\theta}&=\dfrac{y}{a-y}\\ &=\dfrac{a^2-x^2}{a^2+x^2}\hspace{5mm}\color{#3D99F6}\small(1)\\\\ \text{In } &\Delta PDE \text{ we have,}\\\\ PD&=\dfrac{2}{3}a-x\\ PE&=\dfrac{a}{3}\\\\ \cos({90-\theta})&=\frac{PD}{PE}\\\\ \sin\theta&=\dfrac{\dfrac{2}{3}a-x}{\dfrac{a}{3}}\hspace{5mm}\color{#3D99F6}\small(2)\\\\ \text{From }&\color{#3D99F6}\small(1) \color{#333333}\normalsize\text{ and } \color{#3D99F6}\small(2)\\ \dfrac{a^2-x^2}{a^2+x^2}&=\dfrac{\dfrac{2}{3}a-x}{\dfrac{a}{3}}\\ \dfrac{a^2-x^2}{a^2+x^2}&=\dfrac{2a-3x}{a}\\ a(a^2-x^2)&=(a^2+x^2)(2a-3x)\\ a^3-ax^2&=2a^3-3a^2x+2ax^2-3x^3\\ \implies 2x^3&=(a^3-3a^2x+3ax^2-x^3)\\ \implies 2x^3&=(a-x)^3\\ \implies \dfrac{(a-x)^3}{x^3}&=2\\ \implies \dfrac{PB}{PA}&=\dfrac{(a-x)}{x}=\sqrt[3]2\approx\color{#EC7300}\boxed{\color{#333333}1.2599}\end{aligned}$

$\overline{AP}^2=(l-x)^2-x^2=(l-x+x)(l-x-x)=l(l-2x) \implies \overline{AP}=\sqrt{l(l-2x)}$

$\overline{BP}=l-\overline{AP}=l-\sqrt{l(l-2x)}$

Let $l=1$ from now on to simplify the expressions. So far we have:

$\overline{AP}=\sqrt{1-2x}; \qquad \overline{BP}=l-\overline{AP}=1-\sqrt{1-2x}; \qquad \dfrac{\overline{BP}}{\overline{AP}}=\dfrac{1}{\sqrt{1-2x}}-1$

We now find $x$ noting that triangles $\triangle{ACP}$ and $\triangle PRQ$ are similar:

$\dfrac{x}{l-x}=\dfrac{\overline{BP}-l/3}{l/3}$

Being $l=1$ and $\overline{BP}=1-\sqrt{1-2x}$ we obtain $\dfrac{x}{1-x}=3(1-\sqrt{1-2x})-1$

After some algebraic manipulation, we get the following 3rd-degree polynomial on $x$ and calculate its unique real root*:

$18x^3-36x^2+24x-5=0 \implies x=\dfrac{4-\sqrt[3]{4}}{6}\approx 0.4021$

The desired value is $\dfrac{\overline{BP}}{\overline{AP}}=\dfrac{1}{\sqrt{1-\frac{4-\sqrt[3]{4}}{3}}}-1\approx \boxed{1,2599}$

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*It is a laborious task to calculate the root's exact value . I lazily obtained it using Geogebra CAS.