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{\displaystyle {\begin{aligned}&\left(\sum _{n=0}^{p-1}{\text{bit}}_{n}\times 2^{-n}\right)\times 2^{e}\\={}&\left(1\times 2^{-0}+1\times 2^{-1}+0\times 2^{-2}+0\times 2^{-3}+1\times 2^{-4}+\cdots +1\times 2^{-23}\right)\times 2^{1}\\\approx {}&1.5707964\times 2\\\approx {}&3.1415928\end{aligned}}}