My solution : Let $ E \equiv BH \cap CA, F \equiv CH \cap AB, X \equiv AI \cap \odot (ABC), Y \equiv EF \cap BC $ . From $ K \in \odot (AH) $ (well-known) $ \Longrightarrow AK $ is the radical axis of $ \odot (AH), \odot (O) $ , so from $ YE \cdot YF=YB \cdot YC \Longrightarrow Y \in AK \Longrightarrow K(A,M;B,C)=(Y,D;B,C)=-1 $ , hence $ AM $ is A-symmedian of $ \triangle ABC \Longrightarrow N $ is the reflection of $ M $ in $ BC \Longrightarrow N \in \odot (BHC) $ . i.e. $ N $ lie on a fixed circle $ \odot (BHC) $ when $ A $ varies on $ \odot (O) $ From $ \measuredangle BPN=\measuredangle AQN=\measuredangle KAX=\measuredangle KMX=\measuredangle BDN \Longrightarrow B, D, P, N $ are concyclic , so $ \measuredangle BPD=\measuredangle BND=\measuredangle KMB=\measuredangle KAB=\measuredangle AQP \Longrightarrow DP $ is tangent to $ \odot (APQ) $ at $ P $ . Similarly, we can prove $ DQ $ is tangent to $ \odot (APQ) $ at $ Q $ $ \Longrightarrow AD $ is A-symmedian of $ \triangle APQ $ (well-known) , so $ AJ $ is the isogonal conjugate of $ AD $ WRT $ \angle BAC \Longrightarrow AJ $ passes through a fixed point $ O $ when $ A $ varies on $ \odot (O) $ . Q.E.D

In above solution for the part $a$,I have a doubt that when $A$ varies on $(ABC)$,$H$ change position,then ?

Here is my solution for this problem Solution a) Let $G$ $\equiv$ $AK$ $\cap$ $BC$ We define again point $N$ is orthogonal projection of $H$ on $AI$ We have: $\dfrac{MB}{MC}$ = $\dfrac{DB}{DC}$ . $\dfrac{KC}{KB}$ = $\dfrac{GB}{GC}$ . $\dfrac{KC}{KB}$ = $\dfrac{AB}{AC}$ It's easy to see that: $\dfrac{NB}{NC}$ = $\dfrac{AB}{AC}$ = $\dfrac{MB}{MC}$ and $\widehat{BNC}$ = $\widehat{BHC}$ = $\widehat{BMC}$ then: $N$ is reflection of $M$ through $BC$ or $MN$ $\perp$ $BC$ So: $N$ $\in$ ($BHC$) which is reflection of ($O$) through $BC$ b) Since: $A(BCDG)$ = $A(PQHK)$ = $-$ 1 and $AK$ tangents ($APQ$) at $A$, we have: $AH$ is $A$ - symmedian line of $\triangle$ $APQ$ But: $AH$, $AO$ is isogonal conjugate with respect to ($O$) then $AO$ is $A$ - median line of $\triangle$ $APQ$ or $A$, $J$, $O$ are collinear So: $AJ$ passes through fixed point is $O$

Attachments:

Part 1 - Let $F\equiv BH\cap CA$,$E\equiv CH\cap AB$. Now it is well known that $K$ is the miquel point of $BCFE$, let $T\equiv AK\cap BC$, then $T$ is the intersection of $EF$ and $BC$. CLAIM -1 N is the Humpty point of $ABC$.

First observation

Second observation

Part -2 Let $S\equiv AH\cap (APNQ)$. CLAIM-2 $A,J,O$ are collinear, where $O$ is the circumcenter of ($ABC$)

First observation

Second observation

Vietnam TST 2015 P2 wrote: given a fixed circle $(O)$ and two fixed point $B,C$ on it.point A varies on circle $(O)$. let $I$ be the midpoint of $BC$ and $H$ be the orthocenter of $\triangle ABC$. ray $IH$ meet $(O)$ at $K$ ,$AH$ meet $BC$ at $D$ ,$KD$ meet $(O)$ at $M$ .a line pass $M$ and perpendicular to $BC$ meet $AI$ at $N$. a) prove that $N$ varies on a fixed circle. b) a circle pass $N$ and tangent to $AK$ at $A$ cut $AB,AC$ at $P,Q$. let $J$ be the midpoint of $PQ$ .prove that $AJ$ pass through a fixed point. Solution: Let $\Delta H_AH_BH_C$ and $\Delta M_AM_BM_C$ be the orthic and medial triangle WT $\Delta ABC$. Let $\stackrel{\longrightarrow}{M_AH}$ $\cap$ $\odot (ABC)$ $=$ $K$ $\implies$ $\angle AKM_A$ $=$ $90^{\circ}$ $\implies$ $K$ $\in$ $\odot (AHH_BH_C)$ $\implies$ $AK,$ $H_BH_C$, $BC$ concur, say at $X_A$. $$-1=( X_A, H_A ; B,C) \overset{ K}{=} ( A, KH_A ~ \cap ~ \odot (ABC) ; B , C) \implies KH_A ~ \cap ~ \odot (ABC) \in A-\text{symmedian}$$Let $A' \in \odot (ABC)$, such $AA'||BC$ $$-1=( B, C ; M_A , \infty_{BC}) \overset{A'}{=} ( B, C ; A'M_A ~ \cap ~ \odot (ABC), A)$$Hence, $A'M_A$ $\cap$ $\odot (ABC)$ $\equiv$ $KH_A \cap \odot (ABC)$ $\equiv K_A$. Hence, $AM_A$ is the reflection of $K_AM_A$ over $BC$. Therefore, If the perpendicular bisector through $K_A$ to $BC$ intersects $AM_A$ at $A_{HM}$, then, $A_{HM}$ is the reflection of $K_A$ over $BC$. Let $H'$ be the reflection of $H$ over $BC$ $\implies$ $H' \in \odot (ABC)$. Now, $$\angle AHK_A=180^{\circ}-\angle AA'K_A=180^{\circ}-\angle H_AM_AK_A \implies H'H_AM_AK_A \text{ cyclic}$$Hence, $HH_AM_AA_{HM}$ is also cyclic. Hence, by $\sqrt{AH \cdot AH_A}$ inversion $A_{HM}$ lies on $\odot (BHC)$. Note: $\odot (BHC)$ is the reflection of $\odot (ABC)$ over $BC$. But, since, $\odot (ABC)$ and $BC$ are fixed $\implies$ $\odot (BHC)$ is also fixed, hence $A_{HM}$ lies on a fixed circle. Now, $$-1=(X_A, H_A; B ,C) \overset{A}{=} ( A, AH_A ~ \cap ~ \odot (AA_{HM}PQ) ; P ,Q)$$$\implies$ $AH_A$ is isogonal of $AJ$ WRT $\Delta APQ$ $\implies$ $AJ$ passes through fixed point $O$ (circumcenter WRT $\Delta ABC)$