Let $ABCD$ be a rhombus Let  $ABCD$ be  a  rhombus and  $E$  and $F$ be  the  middle of the  sides  $AB$  and $BC$.  If  $P$  is  a point  in  the  interior   of  the  rhombus  such   that  $PE = CE$ and $ AP = AF$  then  $B, P, D $ are collinear points. I  have  no  idea. I  tried using coordinates. I  think is  necessary a construction.
$P$ has the same distance from $A$ and $F$ thus it lies on the perpendicular bisector of $AF$ and for similar reason lies on the perp bisector of $CE$.

It follows that $P$ is on the perpendicular bisector of $AC$ which contains both $B$ and $D$ so $B,P,D$ are collinear.

QED

Edit

It cannot be that $PF=AF$ and $B,P,D$ are collinear. Look at the second picture. The red circle is the locus of the point $P$ having distance $AF$ from $A$, the circle with centre $A$ and radius $AF$. As $P$ must be on the perpendicular bisector of the segment $CE$ it follows that line $BD$ does not contain $P$.
The hint.

Use  reflection $S$ of the rhombus in the line $BD$. 

Thus, $S(P)=P$.