I shall present a trigonometrical solution which many of you may dislike. Here it is. For practicality, let's use some variables: $AE=CF=BK=x$, $BE=AF=y$, $EF=2EP=2PF=2z$ and $\angle BEF=\theta$. By simple angle chasing we get $\angle EFC=240^o-\theta=180^o+60^o-\theta$. From the law of cosines in the triangle $AEF$: $4z^2=x^2+y^2-xy$. (1) From the law of cosines in the triangles $BEF$ and $BCF$: $y^2+4z^2-4yzcos\theta=BF^2=(x+y)^2+x^2-2(x+y)xcos60^o$ $\implies$ $cos\theta=\frac{y-2x}{4z}$. (2) From the law of cosines in the triangle $KEP$, and using (1) and (2): $KP^2=(x+y)^2+z^2-2(x+y)zcos\theta=\frac{9x^2+9xy+3y^2}{4}$. (3) From the law of cosines in the triangles $EFC$ and $EBC$: $x^2+4z^2+4xzcos(60^o-\theta)=EC^2=(x+y)^2+y^2-2(x+y)ycos60^o$ $\implies$ $cos(60^o-\theta)=\frac{2y-x}{4z}$. (4) From the law of cosines in the triangle $PFC$, and using (1) and (4): $PC^2=x^2+z^2+2xzcos(60^o-\theta)=\frac{3x^2+3xy+y^2}{4}$. (5) From the law of cosines in the triangle $KBC$: $KC^2=x^2+(x+y)^2+x(x+y)=3x^2+3xy+y^2$. (6) Finally, from (3), (5) and (6), we get: $KC^2=KP^2+PC^2$ $\implies$ $\triangle KPC$ is a right-angled triangle with $\angle KPC=90^o.$ $\blacksquare$

$EK=AB$, take $Q$ onto $BC$ so that $CQ=CF$, and $R$ so that $AR\parallel BC, AR=AE, \triangle ARE$ is equilateral, $ARQC$ is a parallelogram, so $PR=PC$, while triangles $KER, CBK$ are congruent, s.a.s, hence $KR=KC$ and $KP$ is perpendicular bisector of $CR$, done. Best regards, sunken rock