Assume without loss of generality that $AB > AC$, and we consider the configuration shown below. Our proof can be modified for other configurations.
Let $O$ be the circumcenter of triangle $ABC$. Point $D$ lies on $PO$, which is the perpendicular bisector of segment $AB$. Thus, $ABO$ is an isosceles triangle with $AD = BD$. Likewise, $AEC$ is isosceles with $AE = EC$. Set $x = \angle ABD = \angle BAD$ and $y = \angle CAE = \angle ACE$. (Hence $x + y = \angle BAC$.)
Applying the law of sines to triangles $ABM$ and $ACM$ gives
$$\frac{BM}{\sin x} = \frac{AB}{\sin \angle BMA} \quad \text{and} \quad \frac{CM}{\sin y} = \frac{AC}{\sin \angle CMA}.$$
Dividing the two equations yields
$$\frac{BM}{CM} \cdot \frac{\sin y}{\sin x} = \frac{AB}{AC} \cdot \frac{\sin \angle BMA}{\sin \angle CMA} = \frac{AC}{AB},$$
since $\sin \angle BMA = \sin \angle CMA$ (as they are supplementary angles). Therefore,
$$BM = MC \quad \text{if and only if} \quad \frac{\sin x}{\sin y} = \frac{AC}{AB}. \qquad (*)$$
Applying the law of sines to triangles $ABF$ and $ACF$ gives
$$\frac{AF}{\sin x} = \frac{AB}{\sin \angle AFB} \quad \text{and} \quad \frac{AF}{\sin y} = \frac{AC}{\sin \angle AFC}.$$
Dividing these two equations and noting (*) yields
$$\frac{\sin y}{\sin x} = \frac{AB}{AC} \cdot \frac{\sin \angle AFB}{\sin \angle AFC} \quad \text{or} \quad \sin \angle AFB = \sin \angle AFC.$$
Since $\angle ADF$ is an exterior angle of triangle $ABO$, $\angle EDF = 2x$. Likewise, $\angle DEF = 2y$. Thus $\angle EFD = 180^\circ - 2x - 2y = 180^\circ - 2\angle BAC$. Hence $\angle BFC = 2\angle BAC$ (and so $BOFC$ is cyclic), implying that $\angle AFB + \angle AFC = 360^\circ - 2\angle BAC > 180^\circ$ (since $\angle BAC < 90^\circ$). We must have $\angle AFB = \angle AFC = 180^\circ - \angle BAC$. Since $BOFC$ is cyclic and $BOC$ is isosceles with vertex angle $BOC = 2\angle BAC$, $\angle OFB = \angle OCB = 90^\circ - \angle BAC$. Therefore,
$$\angle AFO = \angle AFB - \angle OFB = 180^\circ - \angle BAC - (90^\circ - \angle BAC) = 90^\circ.$$
Since $\angle APO = \angle AFO = \angle ANO = 90^\circ$, points $A$, $P$, $O$, $F$, $N$ lie on a circle.

Solution 2:

We maintain the notations in the first proof. Let $R$ denote the intersection of lines $MP$ and $CE$. Let $\omega$ denote the circumcircle of $APON$. We claim that $R$ lies on $\omega$. Note that $AEC$ is isosceles, so that $MER$ is also isosceles, and so $ARMC$ is an isosceles trapezoid, hence cyclic. Since $PN \parallel MC$, $ARPN$ is also cyclic.
As in the above proof, we can show that $BFOC$ is cyclic. Then $\angle RFO = \angle OBC = 90^\circ - \angle A$. Moreover, $\angle RAO = \angle RAC - \angle OAC = \angle C - (90^\circ - \angle B) = 90^\circ - \angle A$. Thus $AROF$ is cyclic, so that $ARPOFN$ is cyclic.

Solution 3:

We maintain the notations in the first proof. Assume without loss of generality that $AB > AC$, and we consider the configuration shown below. Our proof can be modified for other configurations.
Since $\angle OPB = \angle OMB = \angle OMC = \angle ONC = 90^\circ$, $BPOM$ and $CNOM$ are cyclic. Let ray $AM$ meet the circumcircles of $BPOM$ and $CNOM$ at $X$ and $Y$ (different from $M$), respectively. We have
$$\angle OXY = \angle OXM = \angle OBM = \angle OBC = \angle BCO = \angle MCO = \angle MYO = \angle XYO;$$
that is, triangle $OXY$ is similar to triangle $OBC$. and the rotation centered at $O$ that takes one to the other. In particular, $\angle XOY = \angle BAC$. Note also that $APON$ is cyclic, implying that $\angle NOD = \angle PAN = \angle BAC$. Hence, $\angle XOY = 2\angle NOD$. Combining the facts that $OX = OY$ and $\angle XOY = 2\angle NOD$, we conclude that the image of $X$ under the reflection through line $OD$ (denoted by $R_{OP}$) is the same as that of $Y$ under the reflection through line $ON$ (denoted by $R_{ON}$). Let $F_1$ denote this common image. Note that the image of line $AYE$ under $R_{ON}$ is line $CF_1E$, and the image of line $ADX$ under $R_{OP}$ is line $BDF_1$. Therefore, $F_1$ lies on lines $CE$ and $BD$; that is, $F_1 = F$.
On the other hand, since $X$ lies on the circumcircle of $MBPO$, $F = F_1$ must lie on its image under $R_{OP}$, which is the circumcircle of $NAPO$.

Solution 4:

Invert the figure about a circle centered at $A$, and let $I_X$ denote the image of the point $X$ under this inversion. Find point $G$ so that $AI_B I_G I_C$ is a parallelogram and let $I_Z$ denote the center of this parallelogram. Note that triangle $BAC$ is similar to triangle $I_C AI_B$ and triangles $BAD$ and $I_D AI_B$. Because $M$ is the midpoint of $AB$ and $I_Z$ is the midpoint of $I_B I_C$, we also have the similarity between the triangles $BAM$ and $I_C AI_Z$. Thus
$$\angle AI_G I_B = \angle I_G AI_C = \angle ZAI_C = \angle MAB = \angle DAB = \angle DBA = \angle AI_D I_B.$$
Hence quadrilateral $AI_BI_DI_G$ is cyclic and, by a similar argument, quadrilateral $AI_CI_EI_G$ is also cyclic. Because the images under the inversion of lines $BDF$ and $CFE$ are circles that intersect in $A$ and $I_F$, it follows that $I_G = I_F$; that is, $G = F$.
Next note that $I_B, I_Z$, and $I_C$ are collinear and are the images of $I_P, I_F$ (or $I_G$), and $I_N$, respectively, under a homothety centered at $A$ and with ratio $1/2$. It follows that $I_P, I_F$ and $I_N$ are collinear, and then that the points $A, P, F$ and $N$ lie on a circle.

Solution 5:

Let $A$ be the origin and denote the complex number of each point by the corresponding lowercase letter. Note that $2n = c$ and $2p = b$. Define point $F_2$ such that
$$f_2 = \frac{2np}{m} = \frac{cp}{m} = \frac{bn}{m};$$
that is
$$\frac{f_2}{c} = \frac{p}{m} \quad \text{and} \quad \frac{f_2}{b} = \frac{n}{m}. \qquad (**)$$
The first equation in $(**)$ implies that triangles $AF_2C$ and $APM$ are similar (and have the same orientation). Hence
$$\angle ACF_2 = \angle AMP = \angle MAC = \angle EAC = \angle ECA;$$
that is, $F_2$ lies on line $EC$. Similarly, the second equation in $(**)$ shows that $F_2$ lies on line $BD$. Thus, $F_2$ lies on both $BD$ and $CE$; that is, $F_2 = F$.
We have established the similarity between triangle $AFC$ (which is $AF_2C$) and triangle $APM$. Thus, triangle $ACM$ is similar to triangle $AFP$ (because of spiral similarities about $A$). Hence $\angle AFP = \angle ACM = \angle ANP$, implying that $ANFP$ is cyclic.