We again prove a sub-problem first and a Lemma . $\textbf{LEMMA: }$$ABC$ be an acute-angles triangle. Let $O$ denote the circumcenter of $\odot(ABC)$ and $E, F$ denote the foot of $B,C-$altitudes repectively. Then $AO \perp EF$. $\textbf{PROOF: }$ $$\measuredangle AXE = \measuredangle XAE + \measuredangle AEX = \measuredangle OAC + \measuredangle AEF = (90^{\circ} - \measuredangle CBA) + \measuredangle CEF = 90^{\circ} - \measuredangle CBA + \measuredangle CBF = 90^{\circ}$$$$\implies AO \perp EF.$$ Now the sub-problem . $\textbf{SUB-PROBLEM: }$$ABC$ be an acute-angled triangle. $E, F$ be the foot of the $B, C-$altitudes respectively, $H$ be its othrocenter, and $O$ be its circumcenter. The perpendicular from $H$ to $EF$ meets the line $\ell$ passing through $A$ and parallel to $BC$ at point $P$, $O' = HP \cap BC$ and let $K = \ell \cap \odot(ABC)$. Prove that, $\qquad\ $ (a) $\{O, O', K\}$ are collinear $\qquad\ $ (b) $O'P = O'K$. $\textbf{PROOF: }$Let $M$ denote the midpoint of $BC$, and let $O''$ denote the reflection of $O'$ over $M$, and $A'$ denote the $A-$antipode. It is well known that $A'$ is also the reflection of $H$ over $M$. Firstly, notice that $AKCB$ is cyclic isoceles trapezoid $\implies$ the perpendicular bisector of $BC$, i.e., $OM$, also perpendicularly bisects $AK$. So, by considering the reflection with respect to the line $OM$, we get, \begin{align*} A &\mapsto K\\ O &\mapsto O\\ O' &\mapsto O''. \end{align*} So, showing $\{O, O', K\}$ are collinear, is equivalent to showing that $\{ A, O, O''\}$ are collinear. So, using phantom points, we restate our current problem for one final last time, $\textbf{SUB-PROBLEM: }$$ABC$ be an acute-angled triangle. Let $O$ be its circumcenter, $H$ its orthocenter, $P = AO \cap BC$, $M$ be the midpoint of $BC$ and $E, F$ be the foot of the $B,C-$altitudes respectively. If $P'$ is the reflection of $P$ over $M$, prove that $HP' \perp EF$. Let $A'$ denote the $A-$antipode. Now, consider the homothety centered at $M$ with scale $-1$. It maps the following, \begin{align*} P &\xleftrightarrow{} P'\\ A' &\xleftrightarrow{} H\\ \implies A'P &\xleftrightarrow{} HP'. \end{align*} So, we get that $HP' \parallel A'P \equiv AO$, and by our initial Lemma as $AO \perp EF \implies HP' \perp EF$ which is what we wanted. And with this, we are done with our part of showing that $\{O, O', K\}$ are collinear. Now for the other part, notice that as $AK \parallel O'O''$ and $OM$ is the perpendicular bisector of both $AK$ and $O'O'' \implies AKO''O'$ is a isoceles trapezoid $\implies O'K = O''A$. Also, as $AP \parallel O'O''$ and $AO'' \parallel PO'$ (proved above in part (a)) $\implies APO'O''$ is a parallelogram $\implies O'P = O''A$. These two in turn finally imply that $O'P = O'K$ and we are done again. $\textbf{SOLUTION: }$ Let $O' = HP \cap BC$ and $O$ denote the center of $\odot(ABC)$, $K = \ell \cap \odot(ABC)$ and $X$ be any arbitrary point on $\ell$ such that $P$ lies between $A$ and $X$ (this point is added for ease in naming angles). Now as $PS$ and $PT$ are angle bisectors of $\angle XPO'$ and $\angle APO'$ we have the following. Firstly, $$\measuredangle O'SP \stackrel{XP \parallel ST}{=} \measuredangle XPS = \measuredangle SPO' \implies O'S = O'P,$$and similarly, $$\measuredangle O'TP \stackrel{AP \parallel ST}{=} \measuredangle APT = \measuredangle TPO' \implies O'T = O'P,$$and so we deduce that $O'S = O'P = O'T \implies O'$ is the center of $\odot(SPT)$. Also notice that by our lemma, we have that $O'P = O'K$, and since $O'$ is the center of $\odot(SPT) \implies K \in \odot(SPT)$. Thus to prove $\{\odot(ABC), \odot(PST)\}$ are tangent to each other, it is sufficient to show that $\overline{O-O'-K}$ are collinear which we have already done in our $\textbf{SUB-PROBLEM}$.

Suppose $M$ is the miquel point of the complete quadrilateral $EFBC$. Now let $AP \cap (PMH)=K$. I claim that $K$ is the point of tangency of $(PST)$ and $(ABC)$. 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clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Claim: $ABCK$ is cyclic. Proof: Redefine $K$ to be $AP \cap (ABC)$, suppose $O$ is the circumcenter of $(ABC)$, $O'$ is the circumcenter of $(HBC)$, $N$ is the midpoint of $BC$, $H'$ is the reflection of $H$ over the midpoint of $BC$ and $X$ be the reflection of $H$ over $BC$. It is well known that $H'$ lies on $(ABC)$ and $H'$ is the antipode of $A$ in $(ABC)$. Now notice that $\angle AMH=90^{\circ}=\angle AMH'$, thus $M-H-N-H'$ are collinear. Notice that $\bigtriangleup BOC \cong \bigtriangleup BO'C$ and the points $O$ and $O'$ lie on the perpendicular bisector of $BC$, thus $ON=NP'$ which means $HO'H'O$ is a parallelogram. Now, $$\angle MHP = \angle MH'A= \angle MKA = \angle MKP$$so $K$ lies on $(PMH)$. $\blacksquare$ Claim: Points $PH \cap BC=Z$, $O$ and $K$ are collinear. Proof: Let $AO \cap BC = Z'$, now $NZ=NZ'$ since $HOH'O'$ is a parallelogram. Thus $K$ and $Z'$ are the reflections of $A$ and $Z$ over $ON$. Thus the lines $AZ'$ and $KZ$ are reflections of each other over $ON$ and thus $K-O-Z$ must be collinear. Since $X$ and $H'$ are reflections of each other as well, and $A-O-Z'-H'$ are collinear, $K-O-Z-X$ are collinear as well. $\blacksquare$ Notice that $\bigtriangleup PNK$ is isosceles as, $$ \angle KPN = \angle KAO = \angle AKO = \angle AKN$$Since $\angle TPS=90^{\circ}$, $PN=TN=SN$ which means $KN=TN=SN$ and thus $\angle TKS = 90^{\circ}$ which means $PTSK$ is cyclic. As noted earlier, the centers $O$, $N$ of $(ABC)$ and $(PST)$ are collinear with $K$ thus indeed, $(PST)$ and $(ABC)$ are tangent at $K$. $\blacksquare$

My solution (briefly): Let $M$ be the intersection of $PH$ with $BC$. Then $M$ is the center of the circle $(PST)$. Further, let $ABCX$ be a parallelogram, then by counting angles it is easy to prove that the angle between the lines $PH$ and $CH$ is equal to the angle between the lines $HX$ and $CH$, whence $P, H, M, X$ are collinear. Let $ABCD$ be an isosceles trapezoid with bases $AD$ and $BC$. Then $BC$ is the median perpendicular to $DX$, so $MD = MX$. Now let $PABK$ be a parallelogram, then $K$ lies on $BC$ and the equality of vectors $\overrightarrow{PK} = \overrightarrow{AB} = \overrightarrow{CX}$ is true, from where $PCXK$ is a parallelogram and $M$ is the intersection point of its diagonals, so $MP = MX$. Then we got that $MS = MT = MP = MX = MD$, from where $SPDTX$ is inscribed! After that, by drawing a tangent in $B$ to $(ABC)$ and counting a few angles, we get that this tangent also touches $(PST)$, from which follows the statement of the problem! I am sorry, in picture point L, not K...

Attachments:

Let $PA$ recuts $(ABC)$ at $D$, $AH,AO$ recuts $(ABC)$ at $H',O'$ then $ADO'H'$ rectangle $DH'$ and $AO$ are symmetric wrt the perpendicular bisector of $BC$ ; let $DH'$ hits $BC$ at $K$; since $H,H'$ are symmetric wrt $BC$ then $DK$ and $KH$ are symmetric wrt $BC$ we deduce $KH\parallel AO$. But $AO\perp EF$ then $K,H,P$ are collinear plus $DOA$ isosceles whence $PKD$ is also isoceles More $\angle PTK=\angle TPD=\angle KPT $ hence $K$ is the circumcenter of the right angled $PST$ besides $KP=KD$ thus $PDST$ is cyclic we know $ (PST),(ABC)$ have a common point $D$ besides the centers $K,O$ and $D$ are collinear so we conclude that $ (PST),(ABC)$ are tangent. Best regards RH HAS

Let $PA$ cut $(ABC)$ at $K$ and $AD$ be the $A$-altitude, we show $K$ lies on $(PST)$, the rest is trivial. Let $H'$ be the reflection of $H$ about $BC$, we claim $KH',PH,BC$ are concurrent. This is true because of angle chase: $\angle AHP=\angle B-\angle C = \angle KH'A$. Since $BC$ is the perpendicular bisector of $HH'$, we are done. Now if $PH\cap BC=R$, then $R$ is trivially the centre of $PST$, and furthermore because $R$ lies on $KH'$ we have $\angle RPK=\angle HRD=\angle H'RD=\angle PKR$, so $RP=RK$ as well.