Answers in order: no such P exists; the two perpendicular bisectors are parallel; no circle passes through three collinear points; and no line cuts a circle in three points. All four are the same fact seen from different sides.
The perpendicular bisectors are parallel. Let A, B, C lie on a line ℓ, with midpoints M of AB and N of BC. The perpendicular bisector of AB is the line through M perpendicular to ℓ; the perpendicular bisector of BC is the line through N perpendicular to ℓ. Two lines perpendicular to the same line are parallel. Since M ≠ N (the points are distinct), these are two distinct parallel lines, so they never meet.
P with PB = PC ⇒ P on the ⊥ bisector of BC
These two lines are parallel and distinct ⇒ no common point ⇒ no such P
No circle through collinear points: a circle needs a centre equidistant from all three, i.e. exactly the point P we have just shown cannot exist.
No line cuts a circle in three points: if a line met a circle at three points, those three points would be collinear and concyclic — impossible by the previous line. A line meets a circle in 0, 1 or at most 2 points.