Ans: A
Since the bearing of $Q$ from $A$ is $\text{S}28^\circ\text{E}$, then $\alpha = 28^\circ$.
In $\Delta APQ$, since the bearings of $P$ and $Q$ from $A$ are $\text{N}42^\circ\text{E}$ and $\text{S}28^\circ\text{E}$, then $\angle PAQ = 110^\circ$. It is given that $PA=QA$, then $\angle APQ = \angle AQP$. Therefore, we have
$\begin{array}{rcl}
\angle AQP & = & \dfrac{1}{2} (180^\circ-110^\circ) \\
\alpha + \beta & = & 35^\circ \\
28^\circ + \beta & = & 35^\circ \\
\beta & = & 7^\circ
\end{array}$
Therefore, the bearing of $P$ from $Q$ is $\text{N} 7^\circ \text{E}$.
