解:$(1)$如图所示,$A_{1}(1,$$4),$$B_{1}(3,$$0),$$C_{1}(4,$$3)$
$(2)$设点$P $的坐标为$(x,$$0),$则$P {B} = |x - (-3)| = |x + 3|$
由题意,得$\triangle P {A}B$的面积为$6$
∵$\triangle P {A}B$以$P {B}$为底边,$A$到$x$轴的距离$4$为高
∴$\frac 12×|x + 3|×4 = 6$
即$2|x + 3| = 6,$解得$x = 0$或$x = -6$
∴点$P $的坐标为$(0,$$0)$或$(-6,$$0)$