Let the mirror rotates through an angle $\alpha$ time ${ }^{\text{'}}{\mathrm{t}}^{\text{'}}\mathrm{\alpha }=\mathrm{\omega t}$ The angle turned by reflected ray = 2$\alpha$ (in clock wise direction). Now the angle
made by reflected ray with vertical will be 60-2$\alpha$ 

$$\begin{gathered} \text{ In }\mathrm{\Delta APB},\mathrm{\angle }\mathrm{P}=60-2\mathrm{\alpha }\text{ and }\cot (60-2\mathrm{\alpha })=\frac{\mathrm{y}}{3} \\ \Rightarrow \mathrm{y}=3\cot (60-2\mathrm{\alpha }) \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dt}}=-3{\mathrm{cosec}}^2(60-2\mathrm{\alpha })\left[-2\frac{\mathrm{d\alpha }}{\mathrm{dt}}\right] \end{gathered}$$
Speed of spot ‘P’ is 
$$\begin{gathered} V=6\omega {\mathrm{cosec}}^2(60-2\alpha )\cdots -\left(1\right) \\ \left[\begin{array}{r}\omega \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=1^0\text{ per second }\\ =\frac{\pi }{180}\mathrm{rad}/\mathrm{s}\end{array}\right] \\ \alpha =\omega \mathrm{t}=\frac{\pi }{180}\times 15=\frac{\pi }{12}\mathrm{rad}/\mathrm{s} \\ \therefore \mathrm{V}=6\times \frac{\mathrm{\pi }}{180}\times {\mathrm{cosec}}^2\left[\frac{\mathrm{\pi }}{3}-2\times \left(\frac{\mathrm{\pi }}{12}\right)\right] \\ =\frac{\mathrm{\pi }}{30}\times {\mathrm{cosec}}^2\left(\frac{\mathrm{\pi }}{6}\right)=\frac{4\mathrm{\pi }}{30}=\frac{2\mathrm{\pi }}{15}\mathrm{m}/\mathrm{s} \end{gathered}$$