∵|2a-b|+（a-4）^2=0.
∴2a-b=0且a-4=0,
解得：a=4,b=8,
S矩形OCAB=32.
当时间T秒时,OP=T-4,
设AP与OC相交于D,∵AC∥PB,
∴ΔDAC∽ΔDPO,
OD/OP=CD/AC,
4(8-CD)=(T-4)*CD,
CD=32/T,∴DQ=2T-CD=(2T^2-32)/T,
∴S阴影=SΔOPD+SΔADQ
=4(T-4)^2/T+4(T^2-16)/T
=8(T^2-4T)/T
=16
T^2-6T=0
T=6或T=0(舍去)
∴P运动时间为6秒.