$解：∵??∠ACB=90°，????BC=2，????∠A=30°??$
$∴??AB=4，????AC=2\sqrt 3??$
$如圖，過點??C??作??CH⊥PP'??于點??H，??連接??PC、????P'C??$
$∵將??△ABC??繞點??C??順時針旋轉(zhuǎn)??120°??得到??△A'B'C??$
$∴??∠PCP'=120°，????CP=CP'??$
$∴??∠CPP'=30°??$
$∵??CH⊥PP'??$
$∴??CH=\frac 12PC??$
$由勾股定理易得??PH=P'H=\frac {\sqrt 3}2PC??$
$∴??PP'=\sqrt 3PC??$
$當(dāng)點??P??與點??A??重合時，??CP??有最大值，即??PP'??有最大值，為??\sqrt 3×2\sqrt 3=6??$
$當(dāng)??PC⊥AB??時，??PC??有最小值，即??PP'??有最小值$
$此時??PC=\frac {AC · BC}{AB}=\sqrt 3??$
$∴??PP'??最小值為??\sqrt 3×\sqrt 3=3??$
$∴線段??PP'??的最大值為??6，??最小值為??3??$
