Solution:

a.
Suppose the points lie in the order $A$, $B$, $C$, $D$. If $P$ lies on $AD$, then the result is trivial, and we have equality if $P$ lies outside the segment $AD$. So suppose $P$ does not lie on $AD$.

Let $M$ be the midpoint of $AD$. Take $P'$ so that $P$, $M$, $P'$ are collinear and $PM = MP'$. Then we wish to prove that $PA + AP' > PB + BP'$. Extend $P'B$ to meet $PA$ at $Q$. Then $P'A + AQ > P'Q$, so $P'A + AP > P'Q + QP$. But $QP + BQ > PB$, so $QP + QP' > PB + PB'$. Hence result.

b.
Let the foot of the perpendicular from $B$, $C$ onto $AD$ be $X$, $Y$ respectively. Suppose that $N$, the midpoint of $XY$, is on the same side of $M$, the midpoint of $AD$, as $D$. Then take $P$ to be a remote point on the line $AD$, the opposite side of $A$ to $D$, so that $A$, $D$, $M$ and $N$ are all on the same side of the line $PAD$ from $P$. Then $PA + PD = 2PM < 2PN \leq PB + PC$.

Contradiction. So we must have $N$ coincide with $M$. But we still have $PA + PD = 2PM = 2PN < PB + PC$, unless both $B$ and $C$ are on the line $AD$. So we must have $B$ and $C$ on the line $AD$ and $AB = CD$. It remains to show that $B$ and $C$ are between $A$ and $D$. Take $P = B$. Then if $C$ is not between $A$ and $D$, we have $PC > PD$ (or $PA$), contradiction.