解法一 由点 $P$ 在平行四边形内，可知$$\lambda>0 , \mu>0,$$由 $AP=\dfrac{\sqrt2}{2}$ 及 $\overrightarrow{AB},\overrightarrow{AD}$ 的模长与夹角均已知，可将 $\overrightarrow{AP}=\lambda\overrightarrow{AB}+\mu\overrightarrow{AD}$ 两边各自做点乘，得$$\lambda^2+2\mu^2+\sqrt2\mu\cdot\lambda=\dfrac12,$$令 $\sqrt2\mu=t$，则 $\lambda^2+t^2+\lambda t=\dfrac12$，整理得$$(\lambda+t)^2-\dfrac12=\lambda t\leqslant\left(\dfrac{\lambda+t}{2}\right)^2,$$解得 $\lambda+t\leqslant\dfrac{\sqrt6}{3}$，当且仅当 $\lambda=t=\sqrt2\mu$ 时，取“=”．
解法二 令 $\overrightarrow{AB}=\overrightarrow{a}$，$\overrightarrow{AD}=\sqrt2\overrightarrow{b}$，则 $\left|\overrightarrow{b}\right|=1,\langle \overrightarrow{a},\overrightarrow{b}\rangle=60^\circ$，如下图延长 $AP$ 交线段 $B'D'$ 于点 $E$，则有$$\overrightarrow{AP}=\lambda\overrightarrow{a}+(\sqrt2\mu)\overrightarrow{b}=m\cdot\overrightarrow{AE},$$整理得$$\overrightarrow{AE}=\dfrac{\lambda}{m}\overrightarrow{a}+\dfrac{\sqrt2\mu}{m}\overrightarrow{b},$$因为 $B',E,D'$ 三点共线，所以 $\dfrac{\lambda}{m}+\dfrac{\sqrt2\mu}{m}=1$，整理得$$\lambda+\sqrt2\mu=m,$$再结合$$m=\dfrac{|AP|}{|AE|}=\dfrac{\frac{\sqrt2}{2}}{|AE|}\leqslant\dfrac{\frac{\sqrt2}{2}}{\frac{\sqrt3}{2}}=\dfrac{\sqrt6}{3},$$因此 $\lambda+\sqrt2\mu$ 的最大值为 $\dfrac{\sqrt6}{3}$．