Cutie pi!

Let the expression be:

$\begin{aligned} P & = \sin{\left(\frac{\pi}{14} \right)} \sin{\left(\frac{3\pi}{14} \right)} \sin{\left(\frac{5\pi}{14} \right)} \sin{\left(\frac{7\pi}{14} \right)} \sin{\left(\frac{9\pi}{14} \right)} \sin{\left(\frac{11\pi}{14} \right)} \sin{\left(\frac{13\pi}{14} \right)} \\ & = \sin{\left(\frac{\pi}{14} \right)} \sin{\left(\frac{3\pi}{14} \right)} \sin{\left(\frac{5\pi}{14} \right)} \sin{\left(\frac{\color{#3D99F6}{\pi}}{\color{#3D99F6}{2}} \right)} \sin{\left(\frac{\color{#3D99F6}{5\pi}}{14} \right)} \sin{\left(\frac{\color{#3D99F6}{3\pi}}{14} \right)} \sin{\left(\frac{\color{#3D99F6}{\pi}}{14} \right)} \\ & = \sin^2{\left(\frac{\pi}{14} \right)} \sin^2{\left(\frac{3\pi}{14} \right)} \sin^2{\left(\frac{5\pi}{14} \right)} \\ & = \cos^2{\left(\frac{3\pi}{7} \right)} \cos^2{\left(\frac{2\pi}{7} \right)} \cos^2{\left(\frac{\pi}{7} \right)} \\ & = \frac{1}{8} \left[ 1+ \cos{\left(\frac{6\pi}{7} \right)} \right] \left[1+ \cos{\left(\frac{4\pi}{7} \right)} \right] \left[ 1+ \cos{\left(\frac{2\pi}{7} \right)} \right] \end{aligned}$

We know that:

$\begin{aligned} \cos{\left(\frac{2\pi}{7} \right)} + \cos{\left(\frac{4\pi}{7} \right)} + \cos{\left(\frac{6\pi}{7} \right)} = - \frac{1}{2} \end{aligned}$

Let $\space a = \cos{\left(\frac{2\pi}{7} \right)}$ , then:

$\begin{aligned} a + 2a^2-1 + 4a^3 - 3a & = - \frac{1}{2} \\ 4a^3 + 2a^2 - 2a - \frac{1}{2} & = 0 \\ 8a^3 + 4a^2 - 4a - 1 & = 0 \end{aligned}$

Now, we have:

$\begin{aligned} P & = \frac{1}{8} \left[ 1+ \cos{\left(\frac{6\pi}{7} \right)} \right] \left[1+ \cos{\left(\frac{4\pi}{7} \right)} \right] \left[ 1+ \cos{\left(\frac{2\pi}{7} \right)} \right] \\ & = \frac{1}{8} \left(1+4a^3-3a \right) \left(1+2a^2-1 \right) \left(1+a \right) \\ & = \frac{1}{8} \left(4a^3-3a + 1 \right) \left(2a^2 \right) \left(a + 1\right) \\ & = \frac{1}{8} \left(8a^3 + 4a^2 - 4a - 1 - 4a^2-2a + 1 + 2 \right) \left(2a^2 \right) \left(a + 1\right) \\ & = \frac{1}{8} \left(- 4a^2-2a + 3 \right) \left(a^2 \right) \left(a + 1\right) \\ & = \frac{1}{16} \left(- 8a^3-4a^2 + 4a +1 -4a -1 + 6a \right) \left(a \right) \left(a + 1\right) \\ & = \frac{1}{16} \left(2a - 1 \right)\left(a \right) \left(a + 1\right) \\ & = \frac{1}{16} \left(2a - 1 \right) \left(a^2 + a \right) \\ & = \frac{1}{16} \left(2a^3 +a^2 -a \right) \\ & = \frac{1}{64} \left(8a^3 +4a^2 -4a -1 + 1 \right) \\ & = \frac{1}{64} \end{aligned}$

$\Rightarrow p + q = 1 + 64 = \boxed{65}$

We can solve this using complex numbers.

Let $w^{14}=1 \Longrightarrow w^k=e^{\frac{2 \pi k}{14}}$ . Now, we have the following identity:

$|1-w^k|=2\sin \left(\dfrac{\pi k}{14}\right)$

But first, note that we can factor the first expression:

$w^{14}-1=0 \Longrightarrow (w^7+1)(w^7-1)=0$

The first factor has roots $w, w^3, w^5, w^7, w^9, w^{11}$ and $w^{13}$ and the second one has roots $w^2, w^4, w^6, w^8, w^{10}, w^{12}$ and $1$ .

Now, the expression we want is:

$\sin{\left(\frac{\pi}{14} \right)} \sin{\left(\frac{3\pi}{14} \right)} \sin{\left(\frac{5\pi}{14} \right)} \sin{\left(\frac{7\pi}{14} \right)} \sin{\left(\frac{9\pi}{14} \right)} \sin{\left(\frac{11\pi}{14} \right)} \sin{\left(\frac{13\pi}{14} \right)}$

But with the identity stated above, that becomes:

$\dfrac{1}{2^7} |1-w||1-w^3||1-w^5||1-w^7||1-w^9||1-w^{11}||1-w^{13}|$

Finally, using $|a||b|=|ab|$ :

$\dfrac{1}{2^7} |(1-w)(1-w^3)(1-w^5)(1-w^7)(1-w^9)(1-w^{11})(1-w^{13})|$

And the fact that, by the fundamental theorem of algebra, these factors represent the roots of $P(x)=x^7+1$ as stated above, then we have:

$\dfrac{1}{2^7} |P(1)|=\dfrac{1}{2^7}|1^7+1|=\boxed{\dfrac{1}{64}}$

So, our final answer is $64+1=\boxed{65}$ .