Let $A_1,A_2,B_1,B_2,C_1,C_2$ denote the circumcenters of $\triangle PBF,$ $\triangle PCE,$ $\triangle PCD,$ $\triangle PAF,$ $\triangle PAE,$ $\triangle PBD,$ respectively. Perpendiculars to $AD,BE,CF$ through its endpoints $A,B,C,D,E,F$ clearly bound a hexagon whose opposite sides are parallel and whose diagonals bisect each other at $O,$ because $O$ is equidistant from its opposite sides. Vertices of this hexagon are nothing but the images of $A_1,A_2,B_1,B_2,C_1,C_2$ under homothety with center $P$ and coeffient $2$ $\Longrightarrow$ segments $A_1A_2,B_1B_2,C_1C_2,OP$ bisect each other at $K.$ This is true for arbitrary $P.$ Now, it suffices to prove that $A_1A_2=B_1B_2=C_1C_2.$ By Stewart theorem for the median $PK$ of $\triangle PA_1A_2,$ we obtain $PK^2=\frac{{PA_1}^2+{PA_2}^2}{2}-\frac{{A_1A_2}^2}{4}.$ But the circumradii $PA_1$ and $PA_2$ verify $\frac{PB}{PA_1}=\frac{PC}{PA_2}=2\sin A=\frac{BC}{R}.$ $PK^2=\frac{R^2}{2BC^2} (PB^2+PC^2)-\frac{{A_1A_2}^2}{4}=\frac{R^2}{2BC^2} \left ( 2PL^2+\frac{BC^2}{2} \right)-\frac{{A_1A_2}^2}{4}$ $\Longrightarrow {A_1A_2}^2=4R^2 \left ( \frac{PL^2}{BC^2}+\frac{1}{4} \right) -4PK^2 \ \ (1)$ By similar reasoning, we get the expressions: ${B_1B_2}^2=4R^2 \left ( \frac{PM^2}{CA^2}+\frac{1}{4} \right) -4PK^2 \ \ (2)$ ${C_1C_2}^2=4R^2 \left ( \frac{PN^2}{AB^2}+\frac{1}{4} \right) -4PK^2 \ \ (3)$ Since $\tfrac{PL}{BC}=\tfrac{PM}{CA}=\tfrac{PN}{AB},$ then from $(1),(2),(3),$ we get $A_1A_2=B_1B_2=C_1C_2$ $\Longrightarrow$ $A_1,A_2,B_1,B_2,C_1,C_2$ lie on a circle with center $K$ the midpoint of $OP.$ Remark: Actually there are two points P that satisfy the problem conditions, namely the Isologic points of the medial triangle MNL, which lie on the Euler line of MNL,ABC.

my solution: more complatated way

Let $O$ be the circumcenter of $\triangle ABC$, $Z$ be the circumcenter of $\triangle AFP$, and $R,Y$ the midpoints of $OP,AO$, $v$ the circumradius of $\triangle ABC$, and $\theta=PM/AC=PL/BC=PN/AB$. Then $\triangle AZP \sim \triangle AOC$, so $\triangle AZO \sim \triangle APC$. Thus $ZY/v =PM/AC =\theta$. Also, using the median formula, we have $ZR^2+OR^2=(ZP^2+OZ^2)/2 =(AZ^2+OZ^2)/2=ZY^2+OY^2$, so that $ZR^2=(v\theta)^2+(v/2)^2-OR^2$. So all of the circumcenters in question lie on a circle with center $R$. Sorry about the diagram. I mislabeled the midpoints $K,L,M$ instead of $L,M,N$.

This beautiful solution belongs to my friend 叶添. $\textbf{Lemma1:}$（李为远） Given a hexagon $ABCDEF$ such that the opposite sides are parallel, let $s_1$ be a segment have direction perpendicular to $AB$ and $DE$ and has length equal to the distance between $AB$ and $CD$, similarly define $s_2, s_3$. If $s_1, s_2, s_3$ can form a triangle, then $ABCDEF$ is inscribed in a circle. $\textbf{Corollary2:}$ Given a hexagon $MNPQST$ such that $MQ, NS, PT$ are collinear at $X$, if the segment $MQ, NS, PT$ can form a triangle, then (a): the circumcenter of $\Delta MXN$, $\Delta NXP$, $\Delta PXQ$, $\Delta QXS$, $\Delta SXT$, $\Delta TXM$ are concyclic. (b): the centroid of $\Delta MPS$ and $\Delta NQT$ coincide. $Proof:$ (a): Consider the line perpendicular to $MQ$ at $M$, similarly we have six lines, consider the hexagon formed by the six lines and use lemma$1$. $\textbf{Lemma3:}$ Given two conic $\Gamma _1$ and $\Gamma _2$ such that they are tangent at two points $A, B$. Let $R$ be a point on $AB$ and $P, Q$ be two points on $\Gamma _1$ such that $PQ$ is tangent to $\Gamma _2$. $PR, QR$ meet $\Gamma _1$ at $P', Q'$ respectively. Then $P'Q'$ is tangent to $\Gamma _2$. $Proof:$ Note that the polar of $R$ with respect to $\Gamma _1$ and $\Gamma _2$ are one line, consider a projective transformation that transform the polar to infinity. $\textbf{Lemma4:}$ Given two circle $C_1, C_2$, let $P$ be their radical center. ($P, C_1, C_2$ are coaxial.) $A, B; C, D$ are collinear and lie on $C_1; C_2$ respectively, $AP, BP; CP, DP$ intersect $C_1; C_2$ again at $A', B'; C', D'$ respectively. Then $A', B', C', D'$ are collinear. $Proof:$ Note that the polar of $P$ with respect to $C_1, C_2$ are one line, consider a projective transformation that transform the polar to infinity. $\textbf{Back to the main problem:}$ $O, H$ denote the circumcenter and orthocenter of $\Delta ABC$, respectively. Let $A_1$ be the intersection point of the line tangent to $(ABC)$ at $A$ and $MN$, similarly define $B_1, C_1$. Note that $P$ lies on the Apollonius circle of $\Delta AMN$ and $A_1$ is the center, so $A_1$ is the radical center of $(ABC), (LMN)$ and $P$ $\Rightarrow $ $P$ is the radical center of $(ABC)$ and $(LMN)$. Let $(LMN)\cap EF=\{D_1, D_2\}$, $X$ lies on $BC$ such that $AX\perp BC$. By lemma$4$ we have $P, D_1, L$ and $P, D_2, X$ are collinear respectively. Let $\Gamma $ be the ellipse with foci $O$ and $H$ and tangent with $(LMN)$, it's easy to see that $BC$ is tangent to $\Gamma $, so by lemma$3$ we have $EF$ is also tangent to $\Gamma $ $\Rightarrow $ $\Gamma $ is a inscribed ellipse of $\Delta DEF$ $\Rightarrow $ $H$ is also the orthocenter of $\Delta DEF$ $\Rightarrow $ the centroids of $\Delta ABC$ and $\Delta DEF$ coincide, then the result is true by lemma$2$. $\Box $

My solution proceeds in three steps. Notations : Relabel points $L, M, N$ as $M_a, M_b, M_c$. Let $H_a, H_b, H_c$ be feet of altitudes from $A, B, C$ in $\Delta ABC$. Let $T_a$ be the intersection between $A$-tangent of $\odot(ABC)$ and $BC$. Define $T_b, T_c$ similarly. Step 1 : We prove that $\angle APT_a=90^{\circ}$, using only $\tfrac{PM_b}{PM_c}=\tfrac{AC}{AB}$. Obviously, $DM_b=0.5AC$ and $DM_c=0.5AB$. So $\tfrac{PM_b}{PM_c}=\tfrac{H_aM_b}{H_aM_c}$, which means $P$ lies on $H_a$-Apollonius circle of $\Delta H_aM_bM_c$. Let the center of this circle be $O_a$. Clearly $O_a\in M_bM_c$ and $H_aO_a$ touches nine-points circle of $\Delta ABC$. Reflecting across $M_bM_c$, we get $AO_a$ touches $\odot(AM_bM_c)$, which also touches $\odot(ABC)$. Hence $O_a$ is the midpoint of $AT_a$ so that Apollonius circle is just circle with diameter $AT_a$, implying the conclusion. Step 2 : Let $T=BT_b\cap CT_c$. We prove that $(PA, PT)$ are isogonal w.r.t. $\angle BPC$. By Desargues Involution Theorem on quadrilateral $BT_bCT_c$ and point $P$, we get an involution swapping $(PA, PT), (PB, PC), (PT_b, PT_c)$. But by Step 1, $\angle BPT_b=\angle CPT_c=90^{\circ}$ so this involution must be isogonality w.r.t. $\angle BPC$ and we are done. Step 3 : We use Step 2 to complete the solution. Let $U, V, W, X, Y, Z$ be the circumcenters of triangles $CPD, DPB, BPF, FPA, APE, EPC$ respectively. Clearly $WX\perp PC, XY\perp PA$ and $YZ\perp PB$. Moreover, radical axis of $\odot(BPF), \odot(CPE)$ is the segment joining $P$ and $BF\cap CE$. So by Pascal's Theorem on $BBFCCE$, we get that this radical axis pass through $T$. Hence $WZ\perp PT$, implying $$\measuredangle(WX, XY)-\measuredangle(WZ, ZY) = \measuredangle(PC, PA) - \measuredangle(PT, PB)$$which equal to zero by Step 2. Hence $\{W, X, Y, Z\}$ are concyclic. Similarly $\{U, V, W, X\}, \{Y, Z, U, V\}$ are concyclic. But there pairwise radical axii are $UV, WX, YZ$ respectively which clearly are not concurrent. Hence these three circles coincides, implying the conclusion.