AB = 2r = 14 cm

PB = 12 cm

$ \Large \angle APB=90 ^{\circ} $(angle in the semicircle)

Let AN = x and NB = (14 - x)

$ \Large \therefore In\triangle APB $

$ \Large AB^{2}=PB^{2}+AP^{2} $

$ \Large (14)^{2}=(12)^{2}+(AP)^{2} $

$ \Large 196=144+(AP)^{2} $

$ \Large (AP)^{2}=196-144 $

$ \Large (AP)^{2}=52 $

$ \Large AP=\sqrt{52} $

$ \Large In\triangle APN $

$ \Large AP^{2}=PN^{2}+AN^{2} $

$ \Large (\sqrt{52})^{2}=x^{2}+PN^{2} $

$ \Large PN^{2}=52-x^{2}...........(i) $

$ \Large In\triangle PNB $

$ \Large PB^{2}=PN^{2}+NB^{2} $

$ \Large (12)^{2}=PN^{2}+(14-x)^{2} $

$ \Large PN^{2}=144-(14-x)^{2}.........(ii) $

$ \Large 52-x^{2}=144-196-x^{2}+28x $

$ \Large 28x=104 $

$ \Large x=\frac{104}{28} $

$ \Large x=\frac{26}{7} $

$ \Large NB=14-x $

$ \Large NB=14-\frac{26}{7} $

$ \Large NB=10\frac{2}{7}cm $