（1）根据题意得$\left\{\begin{array}{l}{k+b=3}\\{-5k+b=-3}\end{array}\right.$，
解得$\left\{\begin{array}{l}{k=1}\\{b=2}\end{array}\right.$，
$\therefore$一次函数解析式为$y=x+2$；
$(2)$把$C\left(a+2,2a+1\right)$代入$y=x+2$得$2a+1=a+2+2$，
解得$a=3$，
即$a$的值为$3$；
$(3)$直线$y=x+2$与$y$轴交于点$D$，如图，则$D\left(0,2\right)$，
设$P\left(0,t\right)$，
$\because S_{\triangle ABD}+S_{\triangle ACD}=S_{\triangle ABD}$，
$\therefore \frac{1}{2}\times |t-2|\times \left(5+1\right)=15$，
解得$t=-3$或$t=7$，
$\therefore P$点坐标为$\left(0,-3\right)$或$\left(0,7\right)$.

（1）根据题意得$\left\{\begin{array}{l}{k+b=3}\\{-5k+b=-3}\end{array}\right.$，
解得$\left\{\begin{array}{l}{k=1}\\{b=2}\end{array}\right.$，
$\therefore$一次函数解析式为$y=x+2$；
$(2)$把$C\left(a+2,2a+1\right)$代入$y=x+2$得$2a+1=a+2+2$，
解得$a=3$，
即$a$的值为$3$；
$(3)$直线$y=x+2$与$y$轴交于点$D$，如图，则$D\left(0,2\right)$，
设$P\left(0,t\right)$，
$\because S_{\triangle ABD}+S_{\triangle ACD}=S_{\triangle ABD}$，
$\therefore \frac{1}{2}\times |t-2|\times \left(5+1\right)=15$，
解得$t=-3$或$t=7$，
$\therefore P$点坐标为$\left(0,-3\right)$或$\left(0,7\right)$.