Ro

Since 4 = 4 log 2 = log 2⁴, rearranging we get
K = 2⁴ × sin π/5 × cos π/10 × sin 3π/5 × cos 3π/10
= 16(sin 36°)(cos 18°)(sin 108°)(cos 54°)
= 16(sin 36°)(sin 72°)(sin 72°)(sin 36°)
= [ 4(sin 36°)(sin 72°) ]²
= [ 4 × (√(5 – 2√5) / (√5 – 1)) × (√(5 + 2√5) / (√5 + 1)) ]²
= [ 4 × (√(5² – (2√5)²) / (√5² – 1²)) ]²
= [ 4 × (√(25 – 20) / (5 – 1)) ]²
= [ 4 × (√5 / 4) ]²
= √5²
= 5

The given sum can be written as

$\displaystyle \log_2 16 + \log_2 ( \prod_{k = 1 }^{k=4} \sin \dfrac{k \pi}{5} )$

There is an identity that I found here that states that

$\displaystyle \prod_{k = 1}^{k =n-1} \sin \dfrac{ k \pi }{n} = \dfrac{ n}{2^{n-1}}$

Therefore the given expression reduces to

$\log_2 16 + \log_2 \dfrac{5}{16} = \log_2 5$

Hence, $K = 5$

$\begin{aligned} S & = \small 4+\log_2 \left(\sin \frac \pi 5\right) + \log_2 \left(\cos \frac \pi {10} \right) +\log_2 \left(\sin \frac {3\pi} 5\right) + \log_2 \left(\cos \frac {3\pi}{10} \right) \\ & = \log_2 \left(16 \blue{\sin \frac \pi 5} \cos \frac \pi {10} \blue{\sin \frac {3\pi}5} \cos \frac {3\pi}{10} \right) & \small \blue{\text{Since }\sin \theta = \cos \left(\frac \pi 2 - \theta \right)} \\ & = \log_2 \left(16 \blue{\cos \frac {3\pi}{10}} \cos \frac \pi {10} \blue{\cos \frac \pi{10}} \cos \frac {3\pi}{10} \right) & \small \blue{\text{and }2 \cos A \cos B = \cos (A- B) + \cos (A+B)} \\ & = \log_2 \left(4 \left(\cos \frac \pi 5 + \cos \frac {2\pi}5 \right)^2 \right) & \small \blue{\text{and }\cos \theta = - \cos (\pi - \theta)} \\ & = \log_2 \left(4 \left(\cos \frac \pi 5 - \cos \frac {3\pi}5 \right)^2 \right) & \small \blue{\text{and }\cos \frac \pi 5 + cos \frac {3\pi}5 = \frac 12} \\ & = \log_2 \left(4 \left(2 \blue{\cos \frac \pi 5} - \frac 12 \right)^2 \right) & \small \blue{\text{and }\cos \frac \pi 5 = \frac {1+ \sqrt 5}4} \\ & = \log_2 \left(4 \left(2 \cdot \blue{\frac {1+\sqrt 5}4} - \frac 12 \right)^2 \right) \\ & = \log_2 5 \end{aligned}$

The required answer $K=\boxed 5$ .