Let $K$ be the midpoint of $BC$. It is well-known that $AP$ passes $K$. Note that $M$ is the Miquel point of the complete quadrilateral $ADBKCP$, so we have $M= \odot(ABK) \cap \odot(CPK)$. Similarly we have $N=\odot(ACK) \cap \odot(BPK)$. Now we have the following claim: Claim. The circle $\odot(MNK)$ is tangent to line $BC$, and furthermore $KA$ is the $K$-symmedian of triangle $KMN$. Note that this claim immediately kills the problem. So it remains to prove the claim. Proof. Invert at $K$. Let us denote the images by an apostrophe. First we have $KB'=KC'$, and $M'=A'B' \cap C'P', N'=A'C'\cap B'P'$. Thus $M'N'$ is parallel to $B'C'$, hence $\odot(KMN)$ is tangent to $BC$. Then $KA'$ is the $K$-median of triangle $KM'N'$, thus $KA$ is the $K$-symmedian of triangle $KMN$. So we are done. $\blacksquare$

Let $K$ be the midpoint of $BC$, $L$ be the midpoint of $AP$. It's easy to find that $\triangle MAP$~$\triangle MBC$ and $\triangle NAP$~$\triangle NCB$, so $\triangle MLP$~$\triangle MKC$ and $\triangle NLP$~$\triangle NKB$. Then we can see that $M,L,N,K$ are concyclic, and $BC$ is tangent to $(MNK)$ at $K$. It is left to prove $\frac {ML}{MK}=\frac {NL}{NK}$, which is quit obvious.

Mathlikerchina wrote: Let $K$ be the midpoint of $BC$, $L$ be the midpoint of $AP$. It's easy to find that $\triangle MAP$~$\triangle MBC$ and $\triangle NAP$~$\triangle NCB$, so $\triangle MLP$~$\triangle MKC$ and $\triangle NLP$~$\triangle NKB$. Then we can see that $M,L,N,K$ are concyclic, and $BC$ is tangent to $(MNK)$ at $K$. It is left to prove $\frac {ML}{MK}=\frac {NL}{NK}$, which is quit obvious. Very Nice!!

Complex Numbers Let $b=-1$ and $c=1$, so, $p=a \cdot k$ for some real $k$, because $0$, $P$ and $A$ are collinear. It's easy to see that $MPC \sim MAB \implies m=\dfrac{ac-pb}{a+c-p-b}=\dfrac{a(k+1)}{a(1-k)+2}$, analogously, $n=-\dfrac{a(k+1)}{a(1-k)-2}$. $(MNO)$ is tangent to $BC$ at O $\iff \dfrac{1-0}{n-0} \div \dfrac{0-m}{m-n} \in \mathbb{R} \iff \dfrac{m-n}{mn} \in \mathbb{R} \iff \dfrac{\dfrac{a(k+1)}{a(1-k)+2}+\dfrac{a(k+1)}{a(1-k)-2}}{-\dfrac{a(k+1)}{a(1-k)+2} \cdot \dfrac{a(k+1)}{a(1-k)-2}} = \dfrac{2a(1-k)}{-a(k+1)} = 2\dfrac{k-1}{k+1} \in \mathbb{R}$. OK! Now, we just have to prove that $\overleftrightarrow{OP}$ is symmedian of $MON \iff \dfrac{\dfrac{m+n}{2}}{m}=\dfrac{n}{p} \iff \dfrac{m+n}{mn} p= \dfrac{4}{a(k+1)}p = \dfrac{4k}{k+1} \in \mathbb{R}$. OK!

Let $K$ be the midpoint of $BC$. Claim $1$: $(MNK)$ is tangent to $BC$. Proof $1$: Invert from $P$. $A'PK'B'N'E'$ is a complete quadrilateral and $C'$ is its miquel point. $D'=C'P\cap (A'B'P)$ and $D'A'M'C'K'PA'$ is a complete quadrilateral where $B'$ is its miquel point. $(PB'C')$ and $(PD'E')$ are tangent at $P$ thus, $D'E'\parallel B'C'$. $A'N'C' \sim A'B'M'\sim PB'C'\sim PE'D'$ and $A'N'B'\sim A'C'M'$ hence \[\frac{N'B'}{C'M'}=\frac{A'B'}{A'M'}=\frac{PB'}{PC'}\]Also since $PBK\sim PK'B'$ and $PCK\sim PK'C',$ we have \[\frac{B'K'}{C'K'}=\frac{B'K'}{BK}.\frac{CK}{C'K'}=\frac{PB'}{PK}.\frac{PK}{PC'}=\frac{PB'}{PC'}=\frac{B'N'}{C'M'}\]Thus, $N'M'\parallel B'C'$. This gives the desired result since $K',B',N'$ and $K',C',M'$ are collinear.$\square$ Claim $2$: $KP$ is $K-$symedian in $MNK\iff $ tangents to $(MNK)$ at $M,N$ intersect on $AP$. Proof $2$: Let $MN\cap AK=S$. We will prove that $\frac{SN}{SM}=\frac{KN^2}{KM^2}$. \[\frac{SN}{\sin NCA}=\frac{SK}{\sin MKC} \ \ \text{and} \ \ \frac{SM}{\sin MBA}=\frac{SK}{\sin NKB}\implies \frac{SN}{SM}=\frac{\sin NKA}{\sin MKA}.\frac{\sin NKB}{\sin MKC}\]Since $NKP\sim NCE$ and $MKP\sim MBD,$ we have \[\frac{NK}{NC}=\frac{KP}{CE} \ \ \text{and} \ \ \frac{MK}{MB}=\frac{KP}{BD}\implies \frac{KN}{KM}=\frac{NC}{BM}.\frac{BD}{CE}=\frac{NC}{BM}.\frac{AB}{AC}\]\[(\frac{NC}{BM}.\frac{AB}{AC})^2=\frac{KN^2}{KM^2}\overset{?}{=}\frac{SN}{SM}=\frac{\sin NKA}{\sin MKA}.\frac{\sin NKB}{\sin MKC}\]\[\frac{\sin NKB}{\sin MKC}=\frac{\sin NAC}{\sin MAB}=\frac{NC.\frac{\sin ANC}{AC}}{MB.\frac{\sin AMB}{AB}}=\frac{NC}{NB}.\frac{AB}{AC}\]Also $NAP\sim NCB$ and $MAP\sim MBC$ hence \[\frac{\sin PKM}{\sin MKC}=\frac{\sin ABM}{\sin MAB}=\frac{MA}{MB}=\frac{AP}{BC}=\frac{NA}{NC}=\frac{\sin NCA}{\sin NAC}=\frac{\sin NKP}{\sin NKB}\implies \frac{\sin NKB}{\sin MKC}=\frac{\sin NKA}{\sin AKM}\]Thus, \[\frac{sin NKA}{\sin MKA}.\frac{\sin NKB}{\sin MKC}=(\frac{\sin NKA}{\sin MKA})^2=(\frac{NC}{BM}.\frac{AB}{AC})^2\]As desired.$\blacksquare$

The diagram of the lemma mentioned in Remark: