Circle and triangle

Note that as $PD,PC $ are tangents, hence, $PDOC$ is cyclic. So, $\angle DPC=180-2x$, where $x= \angle DAC$. Also, $\angle DBC=180-x$, and notice that proving your problem is equivalent to proving $B$ is the orthocenter of $\Delta AK_1K$(stronger even). And so, if it must be true, then $\Delta DPC$ must be the orthic triangle, and hence also, $B $ must be its incenter. Now note $\angle DPC=180-2x$, $\angle DBC=180-x$. And $90+0.5(\angle DPC)=90+0.5(180-2x)=90+90-x=(180-x)=\angle DBC$. So we are closer and just need to prove one angle bisection exists. But $PD=PC$ , and $DO=OC$, hence if we let $PO \cap CD = L$, then $PL$ bisects $\angle DPC$(by angle bisector theorem). So it is enough to force that $B$ is indeed the incenter of $\Delta DPC$. (As in the line in the incenter exists, only one point satisfies $\theta = 90+ 0.5(\alpha)$, and we have shown this condition and also proved $B$ lies in the line in which the incenter lies) So, it is the orthocenter of its extriangle $AK_1K$.

Let $G=AB\cap CD$. By definition, $G$ is in the polar of $P$ with respect to $k$, hence by LaHire Theorem we have that $P$ is in the polar of $G$ with respecto to $k$. By Brocard Theorem we have that $K$ is in the polar of $G$ with respect to $k$. Therefore, $PK$ is the polar of $G$ with respect to $k$, this implies that $PK\perp OK$, i.e. $PK\perp AB$.