椭圆 $\dfrac{x^2}{2}+y^2=1$ 的右焦点为 $A(1,0)$，右准线为直线 $x=2$．过点 $P$ 作直线 $PH$ 与直线 $x=2$ 垂直，垂足为 $H$，则$$\dfrac{PA}{PH}=e=\dfrac{1}{\sqrt{2}},$$故$$\dfrac{PB}{PA}=\sqrt{2}\cdot\dfrac{PB}{PH}=\dfrac{\sqrt{2}}{\sin{\angle{PBH}}},$$显然，当且仅当直线 $PB$ 与椭圆 $\dfrac{x^2}{2}+y^2=1$ 相切时，$\dfrac{PB}{PA}$ 取到最大值．
设点 $P$ 坐标为 $\left(x_0,y_0\right)$，当直线 $PB$ 与椭圆 $\dfrac{x^2}{2}+y^2=1$ 相切时，其方程为$$\dfrac{x_0x}{2}+y_0y=1,$$将点 $B(2,0)$ 代入直线 $PB$ 方程，解得$$\begin{cases}
x_0=1,\\
y_0=\pm\dfrac{\sqrt{2}}{2},
\end{cases}$$此时均有 $\sin{\angle{PBH}}=\dfrac{\sqrt{2}}{\sqrt{3}}$．
综上所述，$\dfrac{PB}{PA}$ 的最大值为 $\sqrt{3}$．