As shown in the figure, there is an isosceles triangle ABC with AB as one side of length 4, $\overline { \mathrm { AC } } = \overline { \mathrm { BC } }$, and $\angle \mathrm { ACB } = \theta$. On the extension of segment AB, a point D is taken such that $\overline { \mathrm { AC } } = \overline { \mathrm { AD } }$, and a point P is taken such that $\overline { \mathrm { AC } } = \overline { \mathrm { AP } }$ and $\angle \mathrm { PAB } = 2 \theta$. Let $S ( \theta )$ be the area of triangle BDP. Find the value of $\lim _ { \theta \rightarrow + 0 } ( \theta \times S ( \theta ) )$. (Here, $0 < \theta < \frac { \pi } { 6 }$) [4 points]
As shown in the figure, there is an isosceles triangle ABC with AB as one side of length 4, $\overline { \mathrm { AC } } = \overline { \mathrm { BC } }$, and $\angle \mathrm { ACB } = \theta$. On the extension of segment AB, a point D is taken such that $\overline { \mathrm { AC } } = \overline { \mathrm { AD } }$, and a point P is taken such that $\overline { \mathrm { AC } } = \overline { \mathrm { AP } }$ and $\angle \mathrm { PAB } = 2 \theta$. Let $S ( \theta )$ be the area of triangle BDP. Find the value of $\lim _ { \theta \rightarrow + 0 } ( \theta \times S ( \theta ) )$. (Here, $0 < \theta < \frac { \pi } { 6 }$) [4 points]