Let $M$ be the midpoint of arc $BC$ of $\odot BOC$ not containing $O$. Let $L$ be the isogonal conjugate of $P$ wrt $\triangle ABC$ and let $K'$ be the reflection of $K$ in line $\overline{MAO}$ Claim: $C,M,K'$ colinear Proof: By construction of $K',L$: $$\measuredangle K'CB=\measuredangle CBL=\measuredangle PBA=2 \measuredangle CBA=\measuredangle COA=\measuredangle CBM=\measuredangle MCB$$ Claim: $K',A,P$ colinear Proof: $$\measuredangle K' AB=\measuredangle CAL= \measuredangle PAB$$ Claim: $BK'=BC$ We already have $$\measuredangle K'CB=2 \measuredangle CBA$$. Also: $$\measuredangle CBK'=\measuredangle LCB=\measuredangle ACP=\measuredangle ACB+\measuredangle BCP$$Then: $$\measuredangle BCP=\measuredangle CBP+ \measuredangle BPC=\measuredangle BOC+\measuredangle CBA=5 \measuredangle ABC$$Hence: $$\measuredangle CBK'=5 \measuredangle ABC+\measuredangle CBA=4 \measuredangle ABC$$so: $$\measuredangle CBK'+2 \measuredangle K'CB=4 \measuredangle ABC+4 \measuredangle CBA=0$$so $\triangle CBK'$ is isosceles with $BC=BK'$ The last two claims show $K \equiv K'$ and the the first claim gives $CK$ bisects $\widehat{BC}$ of $\odot BOC$

there is a tipo let $K'$ be the reflection of $L$ in line $\overline{MAO}$

This problem is innocent-looking... I don't know how to do it without using phantom points. Let $D$ be the arc midpoint of major arc $BC$ of $(BOC)$. Let $l$ be the line through $B$ that's parallel to $AC$, $E = AO\cap l$, $F = CD\cap l$. We will define $K'$ as the intersection between $(CBE)$ and line $CD$. Finally, let $G = (CEB)\cap AC$, $P' = AK \cap l$. The goal is to show that $P'$ lies on the circle $(BOC)$, which will imply $K' = K$ and thus $\angle BCK' = \angle BAC = \angle BEC = \angle BK'C$ (observe that $ABEC$ is a rhombus). First, $\angle CK'E = 180^{\circ} - \angle CBE = 180^{\circ} - \angle ACB = \angle K'CG$, so $GCK'E$ is an isosceles trapezoid, meaning that $GE\parallel CD$. Also, observe that shapes $BAGC \sim BCK'F$, so $\frac{AG}{GC} = \frac{CK'}{K'F} = \frac{AK'}{K'P'}$, so $GK'\parallel CP'$. These imply $K'E\parallel P'D$, which means that $BCP'D$ is cyclic, as desired.

Another solution: Let $K'$ be the intersection point of the tangent of $(ABC)$ at $C$ and $AP$. It is obvious that $CK'$ bisects the major arc $BC$, so it suffices to show that $K\equiv{K'}$ or $BC=BK'$. Let $D$ be the intersection of $BP$ with $(ABC)$ and $E$ the intersection of $CK'$ and $BP$. We have: $$\angle{DPC}=\angle{BPC}=180^\circ-\angle{BOC}=180^\circ-2\angle{A}$$$$\angle{CDP}=\angle{A}$$so $\triangle{CPD}$ is isosceles with $PC=PD$. We will prove that $\triangle{CBK'}\sim{\triangle{DPC}}$ which will yield the desired result. The two triangles have $\angle{BCK'}=\angle{PDC}=\angle{A}$, so we have to show that: $$\frac{CK'}{CB}=\frac{DC}{DP}$$Firstly, note that since $BD\parallel{AC}$, $ABDC$ is an isosceles trapezoid with $DC=AB=AC \Longleftrightarrow \angle{C}=\angle{CBD}=\angle{DCE}$. Combined with $\angle{CDE}=\angle{BCE}=\angle{A}$, we have that triangles $\triangle{BCE}$ and $\triangle{CDE}$ are similar to $\triangle{ABC}$, thus isosceles with $CB=CE$ and $DE=DC=AC$ respectively. Since $PE\parallel{AC}$, $\triangle{K'AC}$ and $\triangle{K'PE}$ are similar, so: $$\frac{K'C}{K'E}=\frac{AC}{PE} \Longleftrightarrow$$$$\frac{K'C}{K'E+K'C}=\frac{AC}{PE+AC} \Longleftrightarrow$$$$\frac{K'C}{CE}=\frac{DC}{PE+DE} \Longleftrightarrow$$$$\frac{K'C}{CB}=\frac{DC}{DP}$$The proof is complete.

Attachments:

Let $M$ be the midpoint of arc $\widehat{BC}$ of $\odot(BOC)$ not containing $O$. Redefine $K$ as a point on line $CM$ such that $BK = BC$. We want to show points $A,K,P$ are collinear. Let $AB,AC$ meet $\odot(BOC)$ again at points $X,Y$. Observe $$ \triangle BKC \text{ and } \triangle XPY \text{ are homothetic } $$since corresponding sides are parallel (just angle chase). This finishes the problem! $\blacksquare$

Let $Q$ be a point on the line segment $AC$ such that $AB=BQ$. ( Which there exist such a point , since $\angle A > \angle C$. ) And suppose that $T$ be the midpoint of arc $\widehat {BC}$ in the circumcircle of triangle $\triangle BOC$ , also $K'$ be the second intersection point of the line $CT$ and circle $\odot BQC$. Now , since the quadrilateral $BQCK'$ is cyclic , one can see that : $$\angle BTC=180-2\angle A \implies \angle BCK'=\angle A=\angle BQA=\angle BK'C \implies BC=BK'$$So for proving that $K \equiv K'$ , it's enough to show that the points $A , K' , P$ are collinear. Now since $BOCP$ is cyclic we have : $$\angle BQK'=\angle BCK'=\angle A \implies \angle CQK'=\angle CPB=180-2\angle A , AC \parallel BP \implies PC \parallel QK'$$So since $\angle QK'B=\angle QCB$ and $QB=QK'=AB$ , by Thales theorem for proving the collinearity , It's enough to show that : $$ \frac{AQ}{QK'} = \frac {AC}{CP} \iff \frac{AQ}{QK'}=\frac{AQ}{AB}=\frac{sin(\angle 2A)}{sin(\angle A)}=\frac{AC}{CP}$$$$ CP.sin(\angle 2A)=BC.sin(\angle ACB)=AC.sin(\angle A) \implies \frac{sin(\angle 2A)}{sin(\angle A)}=\frac{AC}{CP}$$As desired.

This one took me a little too much time than I would have liked. Let $M$ denote the major arc midpoint of $BC$ (i.e. the intersection of the tangents to $(ABC)$ at $B$ and $C$) and $K'$ be the point on $CM$ such that $BC=BK'$. We let $X$ and $Y$ be the second intersections of $AB$ and $AC$ with $(BOC)$. Then, it is clear that due to symmetry, $XBCY$ is an isosceles trapezoid and hence, $BC \parallel XY$. Further, \[\measuredangle PYC = \measuredangle PBC = \measuredangle ACB = \measuredangle MCY \]from which it is clear that $MC \parallel YP$. And finally, \[\measuredangle BXP = \measuredangle BCP = \measuredangle BPC + \measuredangle BCA = \measuredangle BMC + \measuredangle ABC = \measuredangle CBK' + \measuredangle ABC = \measuredangle ABK' \]from which it also follows that $XP \parallel BK'$ as well. Thus, we conclude that $\triangle XPY \sim \triangle BK'C$ are homothetic. Thus, lines $\overline{BX}$, $\overline{CY}$ and $\overline{K'P}$ must concur, the former of which clearly intersect at $A$ and thus, points $A$ , $K'$ and $P$ are collinear, as desired.

Very straightforward and short trigonometric exercise. Let $\angle ABC=\theta$, and $X$ be the arcmidpoint of $BC$. Let $K'$ be the point on $CX$ and such that $BK'=BC$. We now prove $AP$, $BK'$, $CX$ concur. $\angle BAC=\pi-2\theta$, $\angle BOC=2\pi-4\theta$, $\angle BPC=4\theta-\pi$, $\angle PBC=\theta$, $\angle BCP=2\pi-5\theta$, $\angle BCX-\pi-2\theta$. $\angle K'BC=4\theta-\pi$. Now, by trigonometric Ceva and Law of Sines on triangles $ABP$, $ACP$, $BCP$, $\frac{\sin ABP}{\sin PAC}\cdot \frac{\sin(\pi-\theta)}{\sin(\pi-2\theta)}\cdot \frac{\sin(4\theta-\pi)}{\sin(5\theta-\pi)}=\frac{\sin(2\theta)}{\sin(2\pi-4\theta)}\cdot\frac{\sin(2\pi-5\theta)}{\sin(\theta)}\cdot\frac{\sin(\pi-\theta)}{\sin(\pi-2\theta)}\cdot\frac{\sin(4\theta-\pi)}{\sin(5\theta-\pi)}=1$ as wanted.