The Trigonometric Pie

Similar solution as @Saurav Dosi 's, different presentation.

$\begin{aligned} f(\theta) & = \frac {\sin^2 4\theta-\sin^2 \theta}{\sin \theta \sin 5\theta} & \small \color{#3D99F6}{\text{As }\sin^2 A-\sin^2 B = \sin(A+B)\sin(A-B)} \\ & = \frac {\sin 5\theta \sin 3\theta}{\sin \theta \sin 5\theta} \\ & = \frac {\color{#3D99F6}{\sin 3\theta}}{\sin \theta} & \small \color{#3D99F6}{\text{As }\sin 3\theta = 3\sin \theta-4 \sin^3 \theta} \\ & = \frac {\color{#3D99F6}{3\sin \theta-4 \sin^3 \theta}}{\sin \theta} \\ & = 3-4 \sin^2 \theta \\ & = 1+2(\color{#3D99F6}{1-2 \sin^2 \theta}) & \small \color{#3D99F6}{\text{As }\cos 2\theta = 1-2 \sin^2 \theta} \\ & = 1 + 2\color{#3D99F6}{\cos 2 \theta} \end{aligned}$

Therefore,

$\begin{aligned} P & = \prod_{n=0}^8 \left(f(2^{n-1}\theta) - 1 \right) \\ & = \prod_{n=0}^8 \left(1 + 2\cos (2\cdot 2^{n-1} \theta) - 1 \right) \\ & = \prod_{n=0}^8 \left(1 + 2\cos 2^n \theta - 1 \right) \\ & = \prod_{n=0}^8 2\cos 2^n \theta \\ & = 2^9\cos \theta \ \cos 2 \theta \ \cos 3 \theta \ ...\ \cos 8 \theta \\ & = \frac {2^9\color{#3D99F6}{\sin \theta} \ \cos \theta \ \cos 2 \theta \ \cos 4 \theta \ ...\ \cos 256 \theta}{\color{#3D99F6}{\sin \theta}} \\ & = \frac {2^8 \sin 2 \theta \ \cos 2 \theta \ \cos 4 \theta \ \cos 8 \theta \ ...\ \cos 256 \theta}{\sin \theta} \\ & = \frac {2^7 \sin 4 \theta \ \cos 4 \theta \ \cos 8 \theta \ \cos 16 \theta \ ...\ \cos 256 \theta}{\sin \theta} \\ & = \ ... \\ & = \frac {\sin 512 \theta}{\sin \theta} \quad \quad \small \color{#3D99F6}{\text{Putting }\theta=\frac \pi{513}} \\ & = \frac {\color{#3D99F6}{\sin \frac {512 \pi}{513}}}{\sin \frac \pi{513}} \quad \quad \small \color{#3D99F6}{\text{As }\sin (\pi-x) = \sin x} \\ & = \frac {\color{#3D99F6}{\sin \frac \pi{513}}}{\sin \frac \pi{513}} = \boxed{1} \end{aligned}$

Simplifying the given function,

$f\left( \theta \right) =\frac { \sin { (4\theta +\theta )\sin { (4\theta -\theta ) } } }{ \sin { (\theta )\sin { (5\theta ) } } }$ .......................................................................................................( $\sin ^{ 2 }{ (A)-\sin ^{ 2 }{ (B)=\sin { (A+B)\sin { (A-B) } } } }$ )

$f\left( \theta \right) =\frac { \sin { (5\theta )\sin { (3\theta ) } } }{ \sin { (\theta )\sin { (5\theta ) } } }$

$f\left( \theta \right) =\frac { \sin { (3\theta ) } }{ \sin { (\theta ) } }$

$f\left( \theta \right) =\frac { 3\sin { (\theta )-4\sin ^{ 3 }{ (\theta ) } } }{ \sin { (\theta ) } }$ ..........................................................................................................(Triple angle formula)

$f\left( \theta \right) =3-4\sin ^{ 2 }{ (\theta ) }$

$f\left( \theta \right) =1+2(1-2\sin ^{ 2 }{ (\theta )) }$

$f\left( \theta \right) =1+2\cos { (2\theta ) }$ ............................................................................................................(Double angle formula)

Thus,

$f\left( \theta \right) -1=2\cos { (2\theta ) }$

And hence,

$f\left( { 2 }^{ n-1 }(\theta) \right) -1=2\cos { (2\times { 2 }^{ n-1 }(\theta )) }$

That is,

$f\left( { 2 }^{ n-1 }(\theta) \right) -1=2\cos { ({ 2 }^{ n }(\theta )) }$

Now,

$\prod _{ n=0 }^{ 8 }{ (f({ 2 }^{ n-1 }\left( \theta ) \right) -1)=2\cos { (\theta )\times 2\cos { (2\theta )\times .........\times 2\cos { ({ 2 }^{ 8 }\theta ) } } } }$

$\prod _{ n=0 }^{ 8 }{ (f({ 2 }^{ n-1 }\left( \theta ) \right) -1)={ 2 }^{ 9 }(\cos { (\theta )\times \cos { (2\theta )\times .........\times \cos { ({ 2 }^{ 8 }\theta )) } } } }$

$\prod _{ n=0 }^{ 8 }{ (f({ 2 }^{ n-1 }\left( \theta ) \right) -1)={ 2 }^{ 9 }(\frac { \sin { ({ 2 }^{ 9 }\theta ) } }{ { 2 }^{ 9 }\sin { (\theta ) } } ) }$ ..............................................................................( $\prod _{ r=0 }^{ n-1 }{ \cos { ({ 2 }^{ r }\theta )=\frac { \sin { ({ 2 }^{ n }\theta ) } }{ { 2 }^{ n }\sin { (\theta ) } } } }$ )

$\prod _{ n=0 }^{ 8 }{ (f({ 2 }^{ n-1 }\left( \theta ) \right) -1)=\frac { \sin { ({ 2 }^{ 9 }\theta ) } }{ \sin { (\theta ) } } }$

Putting $\theta =\frac { \pi }{ 513 }$ ,

Answer = $\frac { \sin { (\frac { 512\pi }{ 513 } ) } }{ \sin { (\frac { \pi }{ 513 } ) } }$

Answer = $\frac { \sin { (\pi -\frac { \pi }{ 513 } ) } }{ \sin { (\frac { \pi }{ 513 } ) } }$

Answer = $\frac { \sin { (\frac { \pi }{ 513 } ) } }{ \sin { (\frac { \pi }{ 513 } ) } }$ ....................................................................................................................( $\sin { (\pi -\theta )=\sin { (\theta ) } }$ )

Answer = 1