How to show that the product of a Killing tensor and tangent vector is conserved along a geodesic? I understand that a Killing vector $K^{\mu}$ satisfies,
$$ K^{(\mu;\nu)} = 0 $$
I also know that along a geodesic, the quantity 
$$ p_{\mu} K^{\mu}$$
is conserved, where $p_{\mu}$ is the photon 4 momentum, or more generally a tangent vector.
I want to understand why it should be that this quantity is conserved. i.e I want to show that,
$$ \frac{d}{d\lambda} (p_{\mu} K^{\mu}) = 0$$
but cannot see how to even start, given our knowledge of the property of the Killing vector. 
 A: Let's consider a Killing field $K_\mu$. Now the product of the Killing field and the tangent vector is $Q_K = K_\mu \dfrac{dx^\mu}{d\lambda}$.
Now, along a geodesic parametrized by an affine parameter $\lambda$,
$\dfrac{d}{d\lambda}Q_K = \dfrac{d}{d\lambda}\big(K_\mu  \dfrac{dx^\mu}{d\lambda}\big)$
$=\dfrac{\partial K_\mu}{\partial x^\nu} \dfrac{dx^\nu}{d\lambda}\dfrac{dx^\mu}{d\lambda} + K_\mu \dfrac{d^2x^\mu}{d\lambda^2}$
$=K_{\alpha;\nu}+K_\alpha \Gamma^\alpha _{\mu\nu}\dfrac{dx^\nu}{d\lambda}\dfrac{dx^\mu}{d\lambda}+ K_\mu \dfrac{d^2x^\mu}{d\lambda^2} $
$=K_\mu \bigg(\dfrac{d^2x^\mu}{d\lambda^2}+\Gamma^\mu _{\alpha\nu}\dfrac{dx^\alpha}{d\lambda}\dfrac{dx^\nu}{d\lambda}\bigg) = 0$ 
So, the quantity $Q_K$ is conserved along the geodesic. 
Edit The fact that a conserved quantity is apparent along the geodesic is beautifully related to Noether's theorems as illustrated by childofsaturn in this answer.  
A: One can directly show that it's conserved along geodesics as Dvij has done, but it's also illuminating to note that this is just a special case of Noether's theorem. In particular whenever a metric has an isometry generated by a vector field $K^{a}$ the Lagrangian $L = \frac{1}{2}g_{ab}\dot{x}^a \dot{x}^b$ is invariant under the transformation $\delta x^{a} = \epsilon K^{a}$. Now recall that when a Lagrangian is invariant under some transformation $\delta x$, the corresponding Noether charge is $J = \delta x \frac{\partial L}{\partial \dot{x}}$. Here this just becomes $g_{ab}K^{a}\dot{x}^b.$
