15. (12 points) The sequence $\left\{a_{n}\right\}$ satisfies
$$
\begin{array}{l}
a_{1}=\frac{1}{2}, a_{n+1}=a_{n}^{2}+a_{n}(n \in \mathbf{N}), \\
b_{n}=\frac{1}{1+a_{n}}, S_{n}=b_{1}+b_{2}+\cdots+b_{n}, \\
P_{n}=b_{1} b_{2} \cdots b_{n} .
\end{array}
$$

Try to find the value of $2 P_{n}+S_{n}$.
Three, 15. Since $a_{1}=\frac{1}{2}, a_{n+1}=a_{n}^{2}+a_{n}, n \in \mathbf{N}$, therefore, $a_{n+1}=a_{n}\left(a_{n}+1\right)$. Then
$$
\begin{array}{l}
b_{n}=\frac{1}{1+a_{n}}=\frac{a_{n}^{2}}{a_{n} a_{n+1}}=\frac{a_{n+1}-a_{n}}{a_{n} a_{n+1}}=\frac{1}{a_{n}}-\frac{1}{a_{n+1}}, \\
P_{n}=b_{1} b_{2} \cdots b_{n}=\frac{a_{1}}{a_{2}} \cdot \frac{a_{2}}{a_{3}} \cdots \cdots \frac{a_{n}}{a_{n+1}}=\frac{1}{2 a_{n+1}}, \\
S_{n}=b_{1}+b_{2}+\cdots+b_{n} \\
=\left(\frac{1}{a_{1}}-\frac{1}{a_{2}}\right)+\left(\frac{1}{a_{2}}-\frac{1}{a_{3}}\right)+\cdots+\left(\frac{1}{a_{n}}-\frac{1}{a_{n+1}}\right) \\
=2-\frac{1}{a_{n+1}} .
\end{array}
$$

Therefore, $2 P_{n}+S_{n}=\frac{1}{a_{n+1}}+\left(2-\frac{1}{a_{n+1}}\right)=2$.