\[\begin{array}{l} {\text{а)}} \hfill \\ {\text{Пусть }}SE = h. \hfill \\ FE = \frac{1}{2} \cdot BC = 2 \hfill \\ SF = \sqrt {{h^2} + 4} \hfill \\ BE = \frac{1}{2}BD = \sqrt 6 \hfill \\ SB = \sqrt {{h^2} + 6} \hfill \\ AP \cdot SB = SF \cdot AB \Leftrightarrow AP = \frac{{\sqrt {{h^2} + 4} \cdot 2\sqrt 2 }}{{\sqrt {{h^2} + 6} }} \hfill \\ B{P^2} = A{B^2} - A{P^2} = 8 - \frac{{8\left( {{h^2} + 4} \right)}}{{{h^2} + 6}} = \frac{{16}}{{{h^2} + 6}} \Rightarrow BP = \frac{4}{{\sqrt {{h^2} + 6} }} \hfill \\ KE = \frac{1}{2} \cdot AB = \sqrt 2 \hfill \\ SK = \sqrt {{h^2} + 2} \hfill \\ BC \cdot SK = SB \cdot CQ \Leftrightarrow CQ = \frac{{4 \cdot \sqrt {{h^2} + 2} }}{{\sqrt {{h^2} + 6} }} \hfill \\ B{Q^2} = B{C^2} - C{Q^2} = 16 - \frac{{16 \cdot \left( {{h^2} + 2} \right)}}{{{h^2} + 6}} = \frac{{64}}{{{h^2} + 6}} \Rightarrow BQ = \frac{8}{{\sqrt {{h^2} + 6} }} = 2 \cdot BP \Rightarrow \hfill \\ BP = PQ,{\text{ ч}}{\text{.т}}{\text{.д}}{\text{.}} \hfill \\ {\text{б)}} \hfill \\ PK{\text{ - средняя линия треугольника }}CBQ \Rightarrow PK \bot SB \Rightarrow \alpha = \angle APK{\text{ - искомый}}{\text{.}} \hfill \\ SD = 4 \Leftrightarrow \sqrt {{h^2} + 6} = 4 \Leftrightarrow {h^2} = 10 \hfill \\ CQ = \frac{{4 \cdot \sqrt {{h^2} + 2} }}{{\sqrt {{h^2} + 6} }} = 2\sqrt 3 \Rightarrow PK = \sqrt 3 . \hfill \\ AP = \frac{{\sqrt {{h^2} + 4} \cdot 2\sqrt 2 }}{{\sqrt {{h^2} + 6} }} = \sqrt 7 \hfill \\ AK = \sqrt {A{B^2} + B{K^2}} = 2\sqrt 3 \hfill \\ {\text{Теорема косинусов для треугольника }}APK: \hfill \\ A{K^2} = A{P^2} + P{K^2} - 2 \cdot AP \cdot PK \cdot \cos \alpha \Leftrightarrow \hfill \\ \cos \alpha = - \frac{1}{{\sqrt {21} }} \Leftrightarrow \alpha = \arccos \left( { - \frac{1}{{\sqrt {21} }}} \right) = \pi - \arccos \frac{1}{{\sqrt {21} }}. \hfill \\ \end{array}\]

\[\pi - \arccos \frac{1}{{\sqrt {21} }}\]