Let $\alpha = \cos x + i \sin x$, $\beta= \cos y + i \sin y$, and $\gamma = \cos z + i \sin z$. Then $\alpha,\beta,\gamma$ are complex numbers with magnitude $1$. Given condition implies that \begin{align*} p\alpha\beta\gamma &= p\cos(x+y+z) + ip\sin(x+y+z) \\ &= (\cos x + \cos y + \cos z) + i (\sin x + \sin y + \sin z) = \alpha + \beta + \gamma \end{align*}Hence, \[\frac{1}{\alpha\beta} + \frac{1}{\beta\gamma} + \frac{1}{\gamma \alpha} = p\]Considering the real part of the above formula gives the desired result.

$\cos(x+y)+\cos(y+z)+\cos(x+z)= \cos ((x+y+z)-z)+\cos((y+z+x)-x)+\cos((x+z+y)-y)=\cos(x+y+z) ( \cos(x)+\cos(y)+\cos(z))+\sin(x+y+z)(\sin(x)+\sin(y)+\sin(z))=p$