[Solution] Take a point $N$ on $\overparen{M B}$ such that $\overparen{M N}=\overparen{A C}$, and draw $N Q$ perpendicular to $A B$ at $Q$, where the chord $M N=A C$ has length $x$.

Since $\overparen{C M}=\overparen{M B}$, and $\overparen{A C}=\overparen{M N}$, it follows that $\overparen{B N}=\overparen{A M}$,
thus $P Q N M$ is a rectangle, and the length of $P Q$ is $x$.
Also, $A P=Q B$, so the length of $P B=P Q+Q B$ is $x+(x+1)=2 x+1$.

Therefore, the answer is $(E)$.
$20 \cdot 12$ In the circle on the right, $M$ is the midpoint of arc $CAB$, $MP$ is perpendicular to chord $AB$ at
$P$. If the length of chord $AC$ is $x$, and the length of segment $AP$ is $(x+1)$, then the length of segment $PB$ is
(A) $3x+2$.
(B) $3x+1$
(C) $2x+3$.
(D) $2x+2$.
(E) $2x+1$.
(23rd American High School Mathematics Examination, 1972)