Let $ T $ $ \equiv $ $ AB $ $ \cap $ $ DE $. From $ \measuredangle DKB $ $ = $ $ \measuredangle EFA $ and $ BD $ $ \parallel $ $ AE $ we get $ T $ is the exsimilicenter of $ \odot (BDK) $ $ \sim $ $ \odot (AEF) $, so if $ F^* $ is the image of $ K $ under the homothety with center $ T $ that maps $ \odot (BDK) $ $ \mapsto $ $ \odot (AEF) $, then $ \triangle BDK $ and $ \triangle AEF^* $ are homothetic. From $ \measuredangle F^*BK $ $ = $ $ \measuredangle BF^*A $ $ = $ $ \measuredangle DBC $ and $ \measuredangle F^*DK $ $ = $ $ \measuredangle DF^*E $ $ = $ $ \measuredangle CBD $ $ \Longrightarrow $ $ C, $ $ F^* $ are isogonal conjugate WRT $ \triangle BDK $, so $ \measuredangle KF^*A $ $ = $ $ \measuredangle DKC, $ $ \measuredangle KF^*E $ $ = $ $ \measuredangle BKC $ $ \Longrightarrow $ $ F^* $ $ \equiv $ $ F $. Let $ O $ be the circumcenter of $ \triangle AEF $. Since $ \measuredangle BOD $ $ = $ $ \measuredangle AFE $ $ = $ $ \measuredangle BKD $, so notice $ OB $ $ = $ $ OD $ we get $ O $ is the midpoint of arc $ BD $ in $ \odot (BDK) $, hence $ KO $ is the external bisector of $ \angle BKD $ $ \Longrightarrow $ $ C, $ $ F $ are symmetry WRT $ KO $. i.e. $ KC $ $ = $ $ KF $

Attachments:

Note $\angle BKD=\angle AFE=\angle BOD$ so $B,K,O,D$ cyclic and $O$ is the midpoint of arc $BKD$ in $\odot (BKD)$, hence $KO$ is the external angle bisector of $\angle BKD$. Let $BK,DK$ meet $\odot (ABC)$ at $B',D'$ respectively, then $KB'=KD, KD'=KB$. Let $F'$ be the midpoint of arc $B'D'$, by symmetry on $KO$ it follows that quadrilaterals $KD'F'B'$ and $KBCD$ are congruent $\implies KC=KF'$. Also, $F'B'=AB\implies AF'\parallel BB'\implies \angle AF'K=\angle F'KB'=\angle DKC=\angle AFK\implies F\equiv F'$, thus $KC=KF$.

Similar to above, we claim that $F$ is the midpoint of $\widehat{B'D'}.$ Let $C\ne S\equiv CK\cap \odot(O), F\ne R\equiv FK\cap \odot(O).$ Note that $\widehat{SD'}+\widehat{CD}=\angle CKD=\angle AFR=\widehat{AB}+\widehat{BR}\implies \widehat{SD'}=\widehat{BR}\implies KS=KR,$ it follows that $\triangle SFK\cong \triangle RCK$ by ASA, so done. We don't even use $B,K,O,D$ are concyclic.

So here's a solution with motivation that I got yesterday - which is same as leeky's proof. So we start with the pentagon $ABCDE$. As $\angle BKD = \angle AFE = \angle BOD$, $B, K, O, D$ are cyclic. This means that we can determine $K$ only with $A, B, C, D, E$. Now $F$ is a point on the circle $ABCDE$ such that $\angle BKC = \angle KFE$. The locus of such $F$ must be a circle passing through $A$, so it must intersect the circle $ABCDE$ at a point other than $A$. This point will be $F$. Therefore, once we draw $ABCDE$ we can determine $K$ and $F$. Good. Denote $B', D'$ in the same way as leeky did. Now construct $K$ as above. We need to prove $KO \perp CF$. Notice how $KO$ is the external angle bisector of $\angle BKD$. This follows from the fact that $O$ is the midpoint of arc $BD$ - trivial from symmetry. Now this implies $BD' \parallel DB' \perp KO$. As $C$ is the midpoint of arc $BD$, we need $F$ to be the midpoint of arc $B'D'$. So we construct it that way. Denote $F'$ as the midpoint of arc $B'D'$, and now we show $F' \equiv F$. Notice the isosceles trapezoid $ABB'F'$. This gives us $\angle CKD = \angle F'KB = \angle AF'K$. This is enough to claim $F' \equiv F$ by our first paragraph. We are done. $\blacksquare$

Let $B', C', D', F'$ be the intersections when $BK, CK, DK, FK$ is continued to hit the circle again. Note that by problem condition $$\dfrac{1}{2}(\overarc{CD}+\overarc{C'D'})=\angle AFF'=\dfrac{1}{2}(\overarc{AB}+\overarc{BF'})\implies C'D'=BF'$$since $\overarc{AB}=\overarc{CD}$, so $BD'\parallel C'F'$. Similarly, we can obtain $B'D\parallel C'F'$ so $BDB'D'$ is an isosceles trapezoid.. Since $K=DD'\cap BB'$, the perpendicular bisector of $BD'$, which is also that of $B'D$ and $C'F'$, passes through $K$, so $C'K=F'K$ and $\triangle C'F'K$ is isosceles. Finally, since $$\overarc{C'F}=2\angle C'F'K=2\angle F'C'K=\overarc{CF'}$$$CF\parallel C'F'$ so the perpendicular bisector of $CF$ also passes through $K$, so $KC=KF$ done.

It is easy to get by angle chasing that $B,O,K,D$ are concyclic and that $KO\perp CF$ iff $ \widehat{KBD}=\widehat{FAE}$. Take $T$ a point on the circle such that $TA\parallel KB,\ TE\parallel KD$ and take $\{L\}=AB\cap DE$. As $\triangle{BKD}$ and $\triangle{ATE}$ have parallel sides, we get that $L$ is their center of homothety, so $L-K-T$ collinear and $L$ sends $(BKD)$ to $(ATE)$. Let $S$ be the center of $(KBD)$ and let $G$ be the intersection of the bisector of $\widehat{KBD}$ with this circle. The aforementioned homothety sends $G$ to $C$ and $S$ to $O$, whence $\dfrac{LG}{LC}=\dfrac{LS}{LO}$. Taking into account that $S$ is the midpoint of $GO$, we infer $(L,C,G,O)=-1$. As $GK\perp KO$, we obtain that $KG$ is the bisector of $\widehat{LKC}$, or equivalently $\widehat{BKL}=\widehat{CKD}$. Finally, $\widehat{ATK}=\widehat{BKL}=\widehat{CKD}=\widehat{AFK}$, so $T\equiv F$ and the conclusion follows.