Prove the three lines are concurrent. 
Let $O$ be the circumcenter of $\triangle ABC$ with $\angle
A=60^{\circ}$,  $P$ be an arbitary point on the  circumcircle of
   $\triangle BOC$, and $D,E,F$ be the circumcenters of $\triangle
 BPC,\triangle CPA, \triangle APB$ respectively. Prove $AD,BE,CF$ are
   concurrent.

Some intermediate results:


*

*$AD$ bisects $\angle BAC$,and $OD \perp BC$; 

*$O,P$ are isogonal conjugate points of $\triangle DEF$.



 A: Let us denote by $A',B',C$ respectively the intersections of $AD$, $BE$, $CF$ with the sides $BC$, $CA$, $AB$.



We have also denoted by $a_1,a_2;b_1,b_2,c_1,c_2$ the (lengths of the) angles delimited by lines $AA'D$, $BB'E$, $CC'F$ from $\hat A, \hat B,\hat C$.
Then we have the relations:
$$
\begin{aligned}
a_1+a_2 &=\hat A =60^\circ\ ,\\
b_1+c_2 &=180^\circ-\widehat{BPC}=180^\circ-\widehat{BOC}=180^\circ-2\hat A
% =180^\circ-120^\circ\\ &
=60^\circ\ ,\\
b_2+c_1 
&= 180^\circ-a_1-a_2-b_1-c_2 =60^\circ\ .
\end{aligned}
$$
Then working in the circle $(F)=(ABP)$ we have 
$$
\widehat{AFB}
=\overset\frown{APB}
=\overset\frown{AP}
+\overset\frown{PB}
=2b_2+2a_1\ .
$$
So $\widehat{FAB}=90^\circ-\frac 12\widehat{AFB}=90^\circ-a_1-b_2$, giving
$\widehat{FAC}=90^\circ+a_2-b_2$. 
Similarly,
$\widehat{FBC}=(90^\circ-a_1-b_2)+(b_1+b_2)=90^\circ+b_1-a_1$. 
Let us find a formula for the proportion $C'A:C'B$ using (only) the above data.
$$
\begin{aligned}
\frac{C'A}{C'B}
&=
\frac
{\operatorname{Area}AFC}
{\operatorname{Area}BFC}
=
\frac
{AF\cdot AC\cdot \sin \widehat{FAC}}
{BF\cdot BC\cdot \sin \widehat{FBC}}
=
\frac
{AC\cdot \sin (90^\circ+a_2-b_2)}
{BC\cdot \sin (90^\circ+b_1-a_1)}
\ . 
\\[3mm]
&\qquad\text{ Similarly:}
\\[3mm]
\frac{B'A}{B'C}
&=
\frac
{\operatorname{Area}AEB}
{\operatorname{Area}CEB}
=
\frac
{AE\cdot AB\cdot \sin \widehat{EAB}}
{CE\cdot CB\cdot \sin \widehat{ECB}}
=\frac
{AB\cdot \sin (90^\circ+a_1-c_1)}
{CB\cdot \sin (90^\circ+c_2-a_2)}
\ .
\end{aligned}
$$
As already observed in the OP, the point $D$ is on the angle bisector of $\hat A$, which gives the third needed proportion $D'A:D'B$. Putting all together, and with omission of signs,
$$
\begin{aligned}
\frac{B'C}{B'A}\cdot
\frac{C'A}{C'B}\cdot
\frac{A'B}{A'C}
&=
\frac
{BC\cdot \sin (90^\circ+c_2-a_2)}
{AB\cdot \sin (90^\circ+a_1-c_1)}
\cdot
\frac
{AC\cdot \sin (90^\circ+a_2-b_2)}
{BC\cdot \sin (90^\circ+b_1-a_1)}
\cdot
\frac
{AB}
{AC}
\\
&=
\frac
{\sin (90^\circ+c_2-a_2)}
{\sin (90^\circ+a_1-c_1)}
\cdot
\frac
{\sin (90^\circ+a_2-b_2)}
{\sin (90^\circ+b_1-a_1)}
\\
&=1\ ,
\end{aligned}
$$
because we can express all angles in terms of $a_1,b_1,c_1$, and then 
$$
\begin{aligned}
c_2-a_2=(60^\circ-b_1)-(60^\circ-a_1)=a_1-b_1\ ,\\
a_2-b_2=(60^\circ-a_1)-(60^\circ-c_1)=c_1-a_1\ ,
\end{aligned}
$$
(and since the two values $\sin(90\pm ?)$ coincide,) we obtain the cancellations of sine factors.
Now use the reciprocal of the theorem of Ceva to obtain $AA'$, $BB'$, $CC'$ concurrent.
$\square$
A: Another way to prove this is to use projective geometry maps. 
Observe that $E$ and $F$ are on fixed lines (they lie on perpendicular segment bisector for $AC$ and $AB$ respectively, name them $e$ and $f$). So when moving $P$ on circle $BOC$ we see that angle $\angle FDE$ is constant ($=60^{\circ}$), so the transformation $DE\mapsto DF$ is a projective map (induced by rotation around $D$ for $60^{\circ}$) of pencil of lines through $D$ to it self. This one induce new projective map $E\mapsto F$ from $e$ to $f$. But now we have new projective map from a pencil of lines through $B$ to a pencil of lines through $C$: $$\pi: BE\longmapsto CF$$ 
and let $X$ be their intersection point. Since $BC$ goes to it self so $\pi$ is perspective so $X$ describes some line. So we need only two position of $P$ to show that $X$ is on $AD$ and then you arte done.
Now this you could easly check if you put say $P=I$ and $P=O$ and you are done.
