Solution:
Denote the area of $\Gamma$ by $[\Gamma]$. Let $E$ be the intersection of $AC$ and $BD$. From any point $P'$ on $AB$, we construct points $Q', R', S'$ as follows:
Let the line $P'E$ intersect $CD$ at $R'$. Let $\ell$ be the line perpendicular to $P'R'$ at $E$. Let $\ell$ intersect $BC, DA$ at $Q', S'$ respectively.

Define a continuous function $f$ from a point $P'$ on $AB$ to the interval $[0, 1]$ by $f(P') = \frac{[P'Q'R'S']}{[ABCD]}$ where $Q', R', S'$ are constructed as above.

We see that $f(A) = f(B) = 1$. If we can find a point $P'$ on $AB$ such that $f(P') \le \frac{1}{2}$, then by continuity there exists a point $P$ on $AB$ such that $f(P) = \frac{1}{2}$. Then the construction above will yield the points $P, Q, R, S$ as required.

Next, we prove that $f(P') \le \frac{1}{2}$ if $EP'$ bisects $\angle AEB$. We see that, in such case, $EQ', ER', ES'$ bisect $\angle BEC, \angle CED, \angle DEA$ respectively.

Let $a, b, c$ be the lengths of $EA, EB, EC$ respectively. We have that the lengths of $EP', EQ'$ are $\frac{\sqrt{2ab}}{a+b}$, $\frac{\sqrt{2bc}}{b+c}$ respectively. It follows from the AM-GM inequality that
$$(a+b)(b+c)(ab+bc) \ge 8\sqrt{ab}\sqrt{bc}\sqrt{ab \cdot bc} \\ = 8ab \cdot bc,$$
or
$$\begin{aligned} [ABC] &= \frac{1}{2}(ab + bc) \\ &\ge 4\frac{ab}{a+b}\frac{bc}{b+c} = 4[EP'Q'] \end{aligned}$$
Similarly,
$$\begin{aligned} [ABCD] &= \frac{1}{2}([ABC] + [BCD] + [CDA] + [DAB]) \\ &\ge \frac{1}{2}(4[EP'Q'] + 4[EQ'R'] + 4[ER'S'] + 4[ES'P']) \\ &= 2[P'Q'R'S'] \end{aligned}$$
So, $f(P') \le \frac{1}{2}$ as required.