Complex bash with $\Omega$ as the unit circle, so that $$|a|=|c|=|e|=|x|=1$$$$f = -e$$Then since $\angle AXB = \angle CXD = 90^{\circ}$, we have $$\overline{b} = \frac{a+b-x}{ax}$$$$\overline{d} = \frac{c+d-x}{cx}$$And since $ABCD$ is a parallelogram, we have $$b+d = a+c$$Conjugating gives $$\frac{a+b-x}{ax} + \frac{c+d-x}{cx} = \frac1a + \frac1c$$$$ac+bc-cx+ac+ad-ax = cx+ax$$$$ad+bc = 2(ax+cx-ac)$$This is a system of two linear equations in the two unknowns $b,d$. We solve it by a quick elimination: write $$ab+ad = a^2+ac$$and subtract to get $$ab-bc = a^2+3ac-2ax-2cx$$So $$b = \frac{a^2+3ac-2ax-2cx}{a-c}$$$$d = \frac{c^2+3ac-2ax-2cx}{c-a}$$Now let $O,P$ be the circumcenters of $\triangle BXE,\triangle DXF$ respectively. We find \begin{align*} o &= \frac{ex(b\overline{b}-1)}{b-e-x+\overline{b}ex} \\ &= \frac{ex\left[\frac{(a^2+3ac-2ax-2cx)(2a^2+2ac-3ax-cx)}{ax(a-c)^2} - 1\right]}{\frac{a^2+3ac-2ax-2cx}{a-c}-e-x+\frac{e(2a^2+2ac-3ax-cx)}{a(a-c)}} \\ &= \frac{e\left[(a^2+3ac-2ax-2cx)(2a^2+2ac-3ax-cx) - ax(a-c)^2\right]}{(a-c)\left[a(a^2+3ac-2ax-2cx) - a(a-c)(e+x) + e(2a^2+2ac-3ax-cx)\right]} \\ &= \frac{e(2a^4+8a^3c+6a^2c^2-8a^3x-16a^2cx-8ac^2x+6a^2x^2+8acx^2+2c^2x^2)}{(a-c)(a^3+3a^2c-3a^2x-acx+a^2e+3ace-3aex-cex)} \\ &= \frac{2e(a+c)(a-x)(a^2+3ac-3ax-cx)}{(a-c)(a+e)(a^2+3ac-3ax-cx)} \\ &= \frac{2e(a+c)(a-x)}{(a-c)(a+e)} \end{align*}where the large factor of $a^2+3ac-3ax-cx$ (which divides $b-e-x+\overline{b}ex$) must be nonzero since it is given that $E,X,B$ are not collinear. Similarly, we have $$p = \frac{2e(a+c)(c-x)}{(a-c)(c-e)}$$Now $K$ is the reflection of $X$ over line $OP$. Now line $OP$ has equation $$\frac{o-z}{o-p} \in \mathbb{R}$$$$\frac{\frac{2e(a+c)(a-x)}{(a-c)(a+e)} - z}{\frac{2e(a+c)(a-x)}{(a-c)(a+e)} - \frac{2e(a+c)(c-x)}{(a-c)(c-e)}} \in \mathbb{R}$$$$\frac{(c-e)\left[2e(a+c)(a-x) - z(a-c)(a+e)\right]}{2e(a+c)\left[(a-x)(c-e) - (c-x)(a+e)\right]} \in \mathbb{R}$$$$\frac{(c-e)\left[2e(a+c)(a-x) - z(a-c)(a+e)\right]}{2e(a+c)(ax-cx+2ex-ae-ce)} = -\frac{(c-e)\left[2(a+c)(a-x) - x\overline{z}(a-c)(a+e)\right]}{2(a+c)(ax+cx-2ac+ae-ce)}$$$$(ax+cx-2ac+ae-ce)\left[2e(a+c)(a-x) - z(a-c)(a+e)\right] = -(ax-cx+2ex-ae-ce)\left[2e(a+c)(a-x) - ex\overline{z}(a-c)(a+e)\right]$$$$z = \frac{e\left[2(a+c)(a-x)(2ax-2ac+2ex-2ce) - x\overline{z}(a-c)(a+e)(ax-cx+2ex-ae-ce)\right]}{(a-c)(a+e)(ax+cx-2ac+ae-ce)}$$Factoring out $a+e$ gives $$z = -\frac{e\left[4(a+c)(a-x)(c-x) + x\overline{z}(a-c)(ax-cx+2ex-ae-ce)\right]}{(a-c)(ax+cx-2ac+ae-ce)}$$Then we plug in $z=x$ to get $$k = -\frac{e\left[4(a+c)(a-x)(c-x) + (a-c)(ax-cx+2ex-ae-ce)\right]}{(a-c)(ax+cx-2ac+ae-ce)}$$which we expand to $$k = -\frac{e(4a^2c+4ac^2-3a^2x-10acx-3c^2x+4ax^2+4cx^2+2aex-2cex-a^2e+c^2e)}{(a-c)(ax+cx-2ac+ae-ce)}$$We find the vector $$k-x = \frac{-4a^2ce-4ac^2e+2a^2ex+12acex+2c^2ex-4aex^2-4cex^2-2ae^2x+2ce^2x+a^2e^2-c^2e^2-a^2x^2+c^2x^2+2a^2cx-2ac^2x}{(a-c)(ax+cx-2ac+ae-ce)}$$and its conjugate $$\overline{k}-\overline{x} = \frac{-4a^2ce-4ac^2e+2a^2ex+12acex+2c^2ex-4aex^2-4cex^2-2ae^2x+2ce^2x+a^2e^2-c^2e^2-a^2x^2+c^2x^2+2a^2cx-2ac^2x}{ex(a-c)(ax-cx+2ex-ae-ce)}$$Then $$\ell = -\frac{x-k}{x\overline{k}-1} = \frac{k-x}{x(\overline{k}-\overline{x})} = \frac{e(ax-cx+2ex-ae-ce)}{ax+cx-2ac+ae-ce}$$where the giant factor we canceled in the numerator is nonzero because it is given that $k\neq x$. Then \begin{align*} \ell-k &= \frac{e(a-c)(ax-cx+2ex-ae-ce) + e(4a^2c+4ac^2-3a^2x-10acx-3c^2x+4ax^2+4cx^2+2aex-2cex-a^2e+c^2e)}{(a-c)(ax+cx-2ac+ae-ce)} \\ &= \frac{e(4a^2c+4ac^2-2a^2x-12acx-2c^2x+4ax^2+4cx^2+4aex-4cex-2a^2e+2c^2e)}{(a-c)(ax+cx-2ac+ae-ce)} \\ &= \frac{2e(2x-a-c)(ax+cx-2ac+ae-ce)}{(a-c)(ax+cx-2ac+ae-ce)} \\ &= \frac{2e(2x-a-c)}{a-c} \end{align*}Meanwhile, $$a-b = \frac{2(ax+cx-2ac)}{a-c}$$so we see that $|a-b| = |\ell-k|$. $\blacksquare$

bump for a synthetic solution

I’m not a fan of geo, but it’s not a hard one!