Let $D$ the midpoint of $AP$ and let $Q$ the miquelpoint of $BPAC$, now by spiral similarity we get that $BQDM$ is cyclic, furthermore let $AM \cap (ABC)=G$ then note that $\triangle PNA \sim \triangle BGC$ since $\angle PNA=180-\angle BAC=\angle BGC$ and $\angle BCG=\angle PAN$, so now $\angle ABC=\angle AGC=\angle AND=\angle AMP$ thus $(BMP)$ is tangent to $AM$ as desired thus we are done .

Consider an homotethy at $A$ of factor 2 that takes $P$ to $P^\prime$, $B$ to $B^\prime$ and $D$ to $A^\prime$. The problem is now equivalent to prove that $(P^\prime B^\prime A^\prime)$ is tangent to line $AM$ at $A^\prime$. Now, since $(ADB^\prime)$ is tangent to $AC$, $\angle AP'D =\angle DAC=\angle DA^\prime B$ so $BDA^\prime P^\prime$ is cyclic which finishes since $\angle B^\prime A^\prime A=\angle BDA=\angle BP^\prime A$ as desired.

We have that $$AP=AN\cdot\frac{\sin \angle ANP}{\sin \angle APN}=\frac{1}{2}\cdot AM\cdot \frac{\sin \angle BAC}{\sin \angle MAC}=\frac{1}{2}\cdot AM\cdot \frac{\sin\angle ACB\cdot BC/AB}{\sin \angle ACB \cdot CM/AM}=\frac{AM^2}{AB}$$from the rule of sines. This implies the result by PoP.

sami1618 wrote: We have that $$AP=AN\cdot\frac{\sin \angle ANP}{\sin \angle APN}=\frac{1}{2}\cdot AM\cdot \frac{\sin \angle BAC}{\sin \angle MAC}=\frac{1}{2}\cdot AM\cdot \frac{\sin\angle ACB\cdot BC/AB}{\sin \angle ACB \cdot CM/AM}=\frac{AM^2}{AB}$$from the rule of sines. This implies the result by PoP. Good solution

huh Let $D$ be midpoint of $AC$. We claim $NDPM$ is cyclic. $\angle APN=\angle BAM=\angle DMA$. Hence, $\angle ACB=\angle ADN=\angle PMN$ so the tangency is obvious. (Note we used AST and Corresponding Angles).

Let $A_p$ be $A$ - Humpty point of $\triangle ABC$. Note that $\angle{A_pBC} = \angle{NAP}$ and $\angle{A_pCB} = \angle{A_pAC} = \angle{NPA},$ we have $\triangle NAP \sim \triangle A_pBC$. From this, we have $\angle{MNP} = 180^{\circ} - \angle{ANP} = 180^{\circ} - \angle{BA_pC} = \angle{BAC}$ and $\dfrac{NM}{NP} = \dfrac{AN}{NP} = \dfrac{A_pB}{A_pC} = \dfrac{AB}{AC}$. So $\triangle MNP \sim \triangle ABC$. Hence $\angle{NMP} = \angle{ABC} = 180^{\circ} - \angle{PBM}$ or $AM$ tangents $(BMP)$ at $M$

let $A'$ be the reflection of $A$ on $M$ we have that $ABA'C$ is a parallelogram $$\boxed{\textbf{Claim:}\triangle APN\sim\triangle AA'C}$$Proof: since $AC$ is a tangent to the circumcircle of $\triangle APN$ we have that $\measuredangle APN=\measuredangle A'AC$ and since $AB\parallel AA'$ we have that $\measuredangle BAA'=\measuredangle CA'A $ $$\boxed{\triangle APN\sim\triangle AA'C}$$now we have $$\frac{AN}{AP}=\frac{CA'}{AA'}=\frac{\frac{1}{2}AM}{AP}=\frac{AB}{2 AM}\implies AM^2=AP\cdot AB$$and this indicates that $AM$ is a tangent because of the power of a point

A solution I found during contest : Let $(\omega)$ be a circle with center $A$ and radius $0$. Let $f(X) = Pow_{(\omega)}(X) - Pow_{(ANP)}(X)$. By linearity of PoP, $2f(M) = f(B) + f(C)$ Which is equivalent to $2MA^2 - 2MN\cdot MA = AB^2 - BP\cdot BA + AC^2 - AC^2$. $2MA^2 - MA^2 = AB(AB-BP)$. $MA^2 = AB\cdot AP$. Implying the statement by PoP. $\blacksquare$

Let K be the intersection of AM and (ABC). Clearly by angle chasing:$$\boxed{\triangle APN\sim\triangle CBK}$$$PA=AN\cdot BC/KC$ $AB\cdot CM=AM\cdot KC$ So $MA^2 = AB\cdot AP$ implying the statement by PoP. $\blacksquare$