解：過點(diǎn)?$C$?作?$CH⊥AB$?于點(diǎn)?$H$?
由翻折的性質(zhì)可知?$∠AP C=∠QP C$?
∵?$PQ⊥P A，$?∴?$∠APQ=90°$?
∴?$∠AP C=∠QP C=135°$?
∴?$∠BP C+∠QP B=135°$?
∵?$∠QP B=90°，$?∴?$∠BP C=45°$?
∵?$CH⊥AB，$?∴?$∠PHC=90°$?
∴?$∠HCP=∠BP C=45°，$?∴?$CH=PH$?
在?$Rt △ABC$?中，?$AB=\sqrt {AC^2+BC^2}=\sqrt {4^2+3^2}=5$?
∵?$S_{△ABC}=\frac 12\ \mathrm {A}B·CH=\frac 12\ \mathrm {A}C·BC$?
∴?$CH=\frac {12}5，$??$BH=\sqrt {BC^2-CH^2}=\frac 95$?
∴?$P B=PH+BH=\frac {12}5+\frac 95=\frac {21}5$?
