= Solution
The <four-momentum> is $(E/c,p_x,p_y,p_z)$. It transforms by the same <Lorentz boost> as other four-vectors. For an observer moving at $v$ in the positive $x$ direction, its time and longitudinal components transform as
$$
E'=\gamma(E-vp_x),\qquad p_x'=\gamma(p_x-vE/c^2),
\qquad p_y'=p_y,\quad p_z'=p_z.
$$
Substituting the given momentum components proves all three displayed transformation relations.
A photon has zero <rest mass>, so the <relativistic energy-momentum relation> gives $E=cp$ on the positive-energy branch. Lorentz invariance preserves $E^2-c^2p^2=0$, and $E'=\gamma E(1-(v/c)\cos\theta)>0$ for $|v|<c$. Therefore $E'=cp'$. Dividing the transformed longitudinal momentum by $p'=E'/c$ gives the <relativistic aberration> formula
$$
\boxed{\cos\theta'=\frac{\cos\theta-v/c}{1-(v/c)\cos\theta}.}
$$
As $v\to c^-$, for every fixed $\theta\ne0$ the ratio tends to $-1$: photon directions are concentrated along the negative $x$ direction in the observer's frame. The exactly forward photon is exceptional: if $\theta=0$, then $\cos\theta'=1$ for every $v<c$, and its energy tends to zero. For other fixed angles $E'$ diverges. The limiting angular behavior is not uniform near $\theta=0$, and $v=c$ itself is not an admissible inertial observer velocity.
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