My solution is based on long calculation. So I'm giving outline: Let the intersection point be $D$. We'll easily get $BD$ in terms of $a,b,c$. So we'll also get $AD^2$ using cosine rule. We've $PD^2 \cdot AD^2=CD^2 \cdot BD^2.$ Hence we'll get $PD^2.$ Also we've $AP^2-PD^2=AM^2-MD^2.$ Hence we've $AP^2$ Using $AD=AP+PD$ we'll get a equation in terms of $a,b,c$. From here it's easy to show the iff part.

Let $K$ be the antipode of $A$ WRT $\Gamma.$ Since $KC,BH$ are both perpendicular to $AC$ and $KB,CH$ are both perpendicular to $AB,$ then $HCKB$ is parallelogram $\Longrightarrow$ $K$ is the reflection of $H$ about $M.$ Thus, $\angle APM=90^{\circ}$ implies that $M,H,P$ are collinear. If $D \equiv AP \cap BC,$ then $H$ is also the orthocenter of $\triangle ADM.$ Consequently, $OH \in D$ $\Longleftrightarrow$ $OH \perp AM$ $\Longleftrightarrow$ $OH$ is perpendicular bisector of $\overline{AN}$ $\Longleftrightarrow$ $AH=HN.$

In other way.... Let $(M)$ be the circle with diameter $BC$.Now let $\Gamma$ be the circle with diameter $AM$.Then $AP$ is the radical axis of $(O)$ and $\Gamma$. Again $BC$ is the radicla axis of $(O)$ and $(M)$. Now note that $H$ has the sam e power w.r.t $\Gamma$ and $(M)$. So we just need to prove that $O$ lies on the radical axis of $\Gamma$ and $(M)$.But centre of $\Gamma$ lies on $AM$.Let $OH\cap BC=K$. Then $KOH$ is the radical axis of $\Gamma$ and $(M)$. So $OH\perp AN\Longrightarrow AH=HN$. The only if part can be proven in the same way.

The case when $AB<AC$ is similar with $AB>AC$, then suppose that $AB<AC$. Let $HM$ intersect $\Gamma$ at $Q$ and $R$, with $Q$ on the arc $AB$ and $R$ on the arc $BC$. Then we claim that $\angle AQR=90$. To prove this, suppose that $HM$ intersect $AO$ at point $R'$. Note that $AH||OM$, and $AH=2OM$. Then we get that $\Delta R'MO$ is similar with $\Delta R'AH$ with the ratio of similitude 2. Then $R'A=2OA$ and $R',O,A$ are collinear $\rightarrow$ $R'A$ is the diameter of $\Gamma$, which implies that $R'$ lies on $\Gamma$, and finally $R'=R$ (since $H,M,R'$ are collinear too). Then $RA$ is the diameter of $\Gamma$, and our claim was proven. We see that $\angle AQR=90\rightarrow\angle AQM=90$. Then $Q$ lies on a circle with diameter $AM$, and finally the circle with diameter $AM$ cuts $\Gamma$ at $A$ and $Q$, which directly implies that $Q=P$. Now let $AP$ intersect $BC$ at point $X$. Consider the triangle $XAM$. Note that $AH\bot BC\rightarrow AH\bot XM$, and $MH\bot AP\rightarrow MH\bot AX$. Then $H$ is the orthocentre of $\Delta XAM$, and we get that $XH\bot AM$. Note that $AH=HN$ is equivalent with $H$ lies on the perpendicular bisector of $AN$. Since $O$ is obviously lies on the perpendicular bisector of $AN$, we get that the condition is equivalent with $HO\bot AN\Leftrightarrow HO\bot AM$. We have known that $XH\bot AM$. If $HO\bot AM$, then $X,H,O$ are collinear $\Leftrightarrow$ $AP,BC,OH$ meet at a common point. Then we have that $AP,BC,OH$ are concurrent iff $AH=HN$ (Q.E.D)