Let $H$ be the orthocenter of $\triangle ABC.$ $U$ is the midpoint of $\overline{AH}$ lying on $(O)'$ and $T$ is the reflection of $O$ on $P.$ Reflection $A'$ of $A$ on $D'$ lies on image of $(O)'$ under homothety with center $A$ and coefficient 2 $\Longrightarrow$ $A' \in \odot(HBC) \equiv (O_A).$ Since $O_A$ is the reflection of $O$ on $D,$ then $O_AT \parallel DP,$ but $HA' \parallel UD' (\perp DD')$ $\Longrightarrow$ $O_AT$ is the perpendicular bisector of $\overline{HA'}.$ Since $TO'$ is the perpendicular bisector of $\overline{OH},$ then it follows that $T$ is the circumcenter of $\triangle OHA'$ $\Longrightarrow$ $TO=TH=TA'.$ By similar reasoning, we have $TO=TH=TB'=TC'$ $\Longrightarrow$ both $O$ and $H$ lie on $\odot(A'B'C').$ For part b) we rename $P \equiv L,$ as $P$ was already used for part a). Since $\triangle A'BC \sim \triangle D'FE$ are homothetic, it follows that $\tfrac{XB}{XC}=\tfrac{A'C}{A'B}=\tfrac{D'E}{D'F}.$ Thus in the cyclic $BHXC,$ we get $\tfrac{MB}{MC}=\tfrac{HB}{HC} \cdot \tfrac{XB}{XC}=\tfrac{HB}{HC} \cdot \tfrac{D'E}{D'F}.$ Now multiplying the cyclic expressions together yields $\frac{MB}{MC} \cdot \frac{NC}{NA} \cdot \frac{LA}{LB}=\frac{HB}{HC} \cdot \frac{HC}{HA} \cdot \frac{HA}{HB} \cdot \frac{D'E}{D'F} \cdot \frac{E'F}{E'D}\cdot \frac{F'D}{F'E}=\frac{D'E}{D'F} \cdot \frac{E'F}{E'D}\cdot \frac{F'D}{F'E}.$ Since $DD',EE',FF'$ are concurrent cevians of $\triangle DEF,$ the RHS equals 1, thus by the converse of Menelaus' theorem in $\triangle ABC$ we conclude that $M,N,L$ are collinear.

Part a) is very interesting, this is extension for it Let $ABC$ be a triangle with $D,$ $E,$ $F$ are midpoints of $BC,$ $CA,$ $AB$ respectively. $P$ is any point. Lines $PD,$ $PE,$ $PF$ meet circle $(DEF)$ again at $X,$ $Y,$ $Z$ respectively. Let $U,$ $V,$ $W$ be the reflection of $A,$ $B,$ $C$ in $X,$ $Y,$ $Z$ respectively. a) Prove that $(UVW)$ passes through orthocenter $H$ of $ABC.$ b) Let $K$ be the center $(UVW).$ Prove that $P$ is midpoint of $KO$ with $O$ is circumcenter of $ABC.$ More general problem, see https://artofproblemsolving.com/community/c374081h1619334

Attachments:

This extension is a corollary of Hagge circle. Let $Q$ be isogonal conjugate of $P$ in triangle $DEF$ then $ (UVW)$ is the Hagge circle of the image of $Q$ through the homothety centered at centroid $G$ and ratio -2. Part b is a corollary of lemma 4 I posted at here. https://nguyenvanlinh.files.wordpress.com/2011/12/hagge-circles-revisited.pdf See my two proofs for part a at here. https://nguyenvanlinh.files.wordpress.com/2018/03/vn-tst-2018-p13.pdf

I reallly like this problem. part $a$ Let $(DEF)$ be unit circle and $O'=0$. Rename $d,e,f$ to $x,y,z$ respectively. Then $o=x+y+z$ . Rename $D'$ to $D$ (same for others). Calculating we get $d=\tfrac{p-x}{1-\overline{p}x}$ and smilarly for others. It follows that $a'=2d-y-z$. So we need to show that from $PO'=PO$ it follows that $O,A',B',C'$ are concyclic. $PO'=PO$ is equivalent with $$p\sum \overline {x}+\overline {p} \sum x=\sum \overline {x}\sum x$$Call this $*$ . We can add $x+y+z$ and divide by $2$ and preserve that $O,A',B',C'$ are concyclic and this is equivalent with $$\tfrac{e-x-z}{f-x-y} \tfrac{f+z-d-x}{e+y-d-x} \in R$$Simple computation shows that this is equiv. with $$\tfrac{p-\sum x+\overline {p}y(x+z)}{p-\sum x+\overline {p}z(x+y)}\tfrac{(z-x)(p-x-z+\overline{p}xz)}{y-x)(p-x-y+\overline{p}xy)}\in R$$By simplifying we need to show $$p \overline {p} xyz(z+y)+p^2 \sum xy-p \sum x \sum xy+\overline {p^2} x^2z^2 \sum x-\overline (p \sum xy ) (\sum x ) yz+p\overline{p}yz(x+z)(x+y)=0$$and this can easily be simplified via $*$ so we are done. part $b$ Let $P$ be moving on line $l$ trough $O'$. Then $P \mapsto D' \mapsto A' \mapsto X \mapsto M$ is projective so its sufficient to show problem claim in $4$ cases. When $P$ is $O'$ its trivial. Here comes the interesting part. For second and third case take as $P$ one of the intersections of $l$ and Euler circle. Then move $P$ on Euler circle. Again we need $4$ cases. When $P$ is $D,E,F$ its trivial. When $P$ is foot of $A$ altitude its not hard to show problem statement. By this we showed problem claim for $3$ positions of $P$. For case $4$ take $P$ at infinity , and move $P$ on line at infinity. Again $4$ cases and when $P$ is $\infty_{AB},\infty_{AC},\infty_{CB}$ its trivial. And for finish take $P$ as $\infty_{\perp BC}$. This case is trivial as well ( by this we have 6 cases) and with this we are done.

We use complex numbers. Let $(ABC)$ be the unit circle, and define the transformation $\Phi(x)=2x-a-b-c$. Note that $\Phi(O')=O$ and that $\Phi$ maps the 9-point circle to the unit circle. (a) We claim that $2p$ is the center of the circle $(HA'B'C')$. This is true as \[|h-2p|=|a+b+c-2p|=|\Phi(p)|\]\[|a'-2p|=|2d'-a-2p|=|\Phi(d')+b+c-2p|\]However, note that we know $\Phi$ is rigid and $P,D,D'$ are collinear, so $\Phi(p),\Phi(d),\Phi(d')$ are collinear. As $d,d'$ lie on 9-point circle, $\Phi(d)$ and $\Phi(d')$ lie on the unit circle. Note $\Phi(d)=-a$, so by reflections \[\Phi(d)+\Phi(d')=\Phi(p)+\Phi(d)\Phi(d')\overline{\Phi(p)}\]but $\Phi(d)+\Phi(d')-\Phi(p)=\Phi(d')+b+c-2p$ and the result follows that $|h-2p|=|a'-2p|=|\Phi(p)|$. To show $O$ lies on this circle, note that if $PO=PO'$, the homothety at $O$ with factor $2$ takes $P$ to $P'$ and $O'$ to $H$, so $P'H=PO$. Thus $O\in(A'B'C')$. (b) Note that $A'\in(HBC)$. To show this, we simply need to show the distance from $O_{BC}$ the reflection of $O$ in $BC$ to $A'$ is 1. In particular, \[o_{bc}=b+c\implies |a'-o_{bc}|=|2d'-a-b-c|=|\Phi(d')|=1\]Thus, as the center of $(HBC)$ lies on $OD$, $X$ lies on $(HBC)$ as well. Now let $X'$ be the reflection of $X$ over $BC$. We claim that $q=h-2p$ lies on $ax'$. This is true because $q=a+b+c-2p=-\Phi(p)$, $a=-\Phi(d)$, and $x=\sqrt{bc}-\sqrt{bc}-(-bc)\overline a'=bc\overline a'$ (this isn't technically allowed but if we just let the midpoints of arcs be $m_1$ and $m_2$ it is $m_1+m_2-m_1m_2\overline a'$ but $m_1+m_2=0$ and $m_1m_2=-bc$ so we get the same result), implying \[x'=b+c-bc\overline x=b+c-a'=a+b+c-2d'=-\Phi(d')\]so thus $a,x',q$ are collinear. We never cared about $q$; if we define $y'$ and $z'$ similarly, we know $ax', by', cz'$ concur at a point $Q$. Rename $X'\to X$ and similarly with $Y'$ and $Z'$. Then, by complex intersection (as $m=(\text{reflection of }H\text{ over }BC)X\cap BC=(-bc/a)x\cap bc$) \[m=\frac{-bc+ax+bx+cx}{a+x}=h-\frac{ah+bc}{a+x}=h-\frac{(a+b)(a+c)}{a+x}\]so $MB=\frac{|a+c||b-x|}{|a+x|}$. Thus, by Menelaus, it suffices to show $BX\cdot CY\cdot AZ=CX\cdot AY\cdot BZ$, but this follows by writing $BX=BQ\frac{\sin BQX}{\sin C}$ and the cyclic variations.