Nice Solution: Since $ BCKE$ is harmonic quadrilateral,so if we consider the pencil $ A(BCKE)$ and its intersection with $ \overline{CMN}$,we will obtain that $ (CN,M\infty) = - 1$,hence $ CM = MN$ and we are done. BTW:As you can see from my solution,the property of $ ABCD$ is trapezium is not necessary,it is still correct if $ ABCD$ is just a cyclic quadrilateral

Any solution without projective?

We use complex numbers. Let $(ABC)$ be the unit circle. Since $ABCD$ is a trapezoid, $ad=bc\implies d=\frac{bc}a$ and $p=\frac{(a+d)c^2-2acd}{c^2-ad}=\frac{a^2c^2+bc^3-2abc^2}{ac(c-b)}=\frac{c(a^2-2ab+bc)}{a(c-b)}=\frac{2ce}{c+e}\\\implies (a^2-2ab+bc)c=e(2ac-2ab-a^2+2ab-bc)\implies e=\frac{c(a^2-2ab+bc)}{2ac-a^2-bc}$. Additionally, $\frac{b-p}{b-k}=\frac{\bar b-\bar p}{\bar b-\bar k}=\frac{k(1-b\bar p)}{k-b}\implies k=\frac{p-b}{1-b\bar p}=\frac{\frac{c(a^2-2ab+bc)}{a(c-b)}-b}{1-b\frac{bc-2ac+a^2}{ac(b-c)}}=\frac{c(c(a^2-2ab+bc)+ab(b-c))}{ac(c-b)+b(bc-2ac+a^2)}=\frac{c(ab^2+bc^2+ca^2-3abc)}{a^2b+b^2c+c^2a-3abc}$. For $N$, we have $\frac{a-n}{a-e}=\frac{\bar a-\bar n}{\bar a-\bar e}=\frac{e(1-a\bar n)}{e-a}\implies n+ae\bar n=a+e$ and $\frac{c-n}{a-b}=\frac{\bar c-\bar n}{\bar a-\bar b}=\frac{ab(\bar c-\bar n)}{b-a}\implies n+ab\bar n=ab\bar c+c$. By Cramer's rule, $n=\frac{ab-e(ab\bar c+c-b)}{b-e}$. Similarly, $m=\frac{ab-k(ab\bar c+c-b)}{b-k}$. $n=\frac{ab-\frac{c(a^2-2ab+bc)}{2ac-a^2-bc}(ab\bar c+c-b)}{b-\frac{c(a^2-2ab+bc)}{2ac-a^2-bc}}=\frac{ab(2ac-a^2-bc)-c(a^2-2ab+bc)(ab\bar c+c-b)}{b(2ac-a^2-bc)-c(a^2-2ab+bc)}\\=\frac{3a^2bc-a^3b-a^3b+2a^2b^2-a^2c^2+2abc^2-bc^3-4ab^2c+b^2c^2}{4abc-a^2b-b^2c-a^2c-bc^2}\\ m=\frac{ab-\frac{c(ab^2+bc^2+ca^2-3abc)}{a^2b+b^2c+c^2a-3abc}(ab\bar c+c-b)}{b-\frac{c(ab^2+bc^2+ca^2-3abc)}{a^2b+b^2c+c^2a-3abc}}\\=\frac{a^3b^2-a^3bc-a^2b^3+2a^2bc^2-a^2c^3+2ab^3c-5ab^2c^2+3abc^3+b^2c^3-bc^4}{b^3c+b^2a^2-4b^2ac+4bac^2-bc^3-a^2c^2}\\=\frac{(c-b)(3abc^2-bc^3-a^2c^2-2ab^2c+a^2bc+a^2b^2-a^3b)}{(c-b)(4abc-a^2b-b^2c-a^2c-bc^2)}$ Indeed, $c+n=\frac{3a^2bc-a^3b-a^3b+2a^2b^2-a^2c^2+2abc^2-bc^3-4ab^2c+b^2c^2+c(4abc-a^2b-b^2c-a^2c-bc^2)}{4abc-a^2b-b^2c-a^2c-bc^2}\\=\frac{-2a^3b+2a^2b^2+2a^2bc-2a^2c^2-4ab^2c+6abc^2-2bc^3}{4abc-a^2b-b^2c-a^2c-bc^2}=2m,$ as desired.