Restricting to the case where $P$ is triangular, if the angles of the triangle are $A$, $B$, $C$ and you fold the triangle along the bisector to the angle $C$, then assuming wlog that $A<B$, you obtain an overlapping area whose ratio to that of the full triangle is:

$$r(A,B)=\frac{\sin A}{\sin A + \sin B}$$

Choosing to fold along the bisector of the angle represents a local maximum for the overlapping area, among all choices of folding lines that pass through the same vertex.

You can then choose which vertex to bisect, of the three, as the one that yields the largest value of $r(A,B)$:

$$r_{max}=\max\{r(A,B),r(B,\pi-A-B),r(\pi-A-B,A)\}$$

A lower bound for $r_{max}$ occurs when two of the vertex choices give the same result, and this lower bound gets smaller in the limit that the two angles $A, B$ are both small. Set $B = f A$ and express the equality of two solutions with the function:

$$P(f,A) = r(A, f A)-r(f A, \pi-A-f A)\\ = \frac{\sin A}{\sin A + \sin (f A)} - \frac{\sin (f A)}{\sin (f A) + \sin (\pi-A-f A)}$$

To first order in $A$, this becomes:

$$P(f,A) \approx \frac{1}{1+f} - \frac{f}{1+2f}$$

This becomes zero (i.e. the two ratios for the different choice of vertex match) when:

$$f = \Phi= \frac{1+\sqrt{5}}{2}$$

So $B/A$ is equal to the golden ratio. Inserting this value for $f$ back into either area ratio gives a lower bound of:

$$r_{min}=\frac{1}{1+\Phi}\approx 0.381966$$

**Edited to add**:

A slight improvement can be achieved by allowing the additional possibility of folding along the perpendicular bisector to any of the sides. If the two angles at the endpoints of the side whose bisector is used are $A$ and $B$ and $A \lt B$, then the area ratio is:

$$r_{p}(A,B)=\frac{1}{4} (1 + \tan A \cot B)$$

An approach that included *only* these perpendicular bisectors would do terribly, but adding three more possibilities due to $r_p$ to the original three used to calculate $r_{max}$ yields a slight improvement.

In this case, the lower bound is found by setting:

$$P(f,A) = r_p(A, f A)-r(f A, \pi-A-f A)\\ = \frac{1}{4} (1 + \tan A \cot f A) - \frac{\sin (f A)}{\sin (f A) + \sin (\pi-A-f A)}$$

To first order in $A$, this becomes:

$$P(f,A) \approx \frac{1+f}{4 f} - \frac{f}{1+2f}$$

This is zero when:

$$f = \frac{3+\sqrt{17}}{4} \approx 1.78078$$

and the ratios here are:

$$r_{min} = \frac{\sqrt{17}-1}{8} \approx 0.390388$$

**Edited to add**:

There is a further improvement possible. While using the bisector of an *angle* as a folding line yields a local maximum for the overlapping area (among all lines passing through a given vertex), using the perpendicular bisector of a *side* is *not* a local maximum (among all lines perpendicular to that side).

When the folding line is displaced from the midpoint of the side to lie further away from the smaller of the angles incident on that side, the portion of the original triangle on the small-angle side of the folding line will, after folding, have a non-overlapping piece that needs to be excised, leading to a **quadrilateral** overlapping region. But the increasing base and altitude of the triangle more than compensate for the excision, up until a scaled position for the folding line of:

$$\alpha(A,B) = \frac{2}{3+\tan A \cot B}$$

where as usual we assume $A\lt B$, and $\alpha$ ranges from $0$ to $1$ as the point where the folding line crosses the sides moves from angle $A$ to angle $B$. At $\alpha(A,B)$, the ratio for the overlapping area has its peak of:

$$r_{dp}(A,B) = 1 - \frac{2}{3+\tan A \cot B}$$

In principle, we might worry that this optimal folding line lies past the vertex of the original triangle, complicating the geometry, but that turns out to be impossible. So one of these **displaced-perpendicular** folding lines can be used whenever the $r_{dp}$ for a side gives a better result than all the angle-bisectors.

The new lower bound can be found by the same kind of methods used previously, in the limit where angles $A$ and $B$ both get smaller but maintain a fixed ratio, $B = f A$. Using $r_{dp}$ rather than $r_p$ in the previous calculations, we get:

$$f = 1 + \sqrt{2}$$

and a new lower bound of:

$$r_{min} = \sqrt{2} - 1 \approx 0.414214$$

The figure below shows how the three approaches described (folding along angle bisectors, along perpendicular bisectors to sides, and along the optimal displaced perpendicular) can allow successively greater overlapping areas, for one particular triangle. But for *other* triangles, folding along the angle bisectors will give the best result.

thinnerslices of bread? (I know, I know). $\endgroup$ – Mateusz Kwaśnicki Jun 21 '18 at 9:55