${\mathop{\rm mg}\nolimits} ({\rm{PA}}) = \frac{1}{2}{\rm{m}}{{\rm{v}}^2}{\rm{;mgh}} = \frac{1}{2}{\rm{m}}{{\rm{v}}^2}$

$v^{2}=2 g(P A)=2 g(O A-O P)$

${v^2} = 2g(l - l\cos \theta )$

${T-m g=\frac{m v^{2}}{l}} $

${T-m g=\frac{m}{l} \cdot 2 g((1-\cos \theta)}$

${T-m g=2 m g(1-\cos \theta)}$       ......$(i)$

At $A, \quad T=T_{\max }=2 m g$

On putting this value of $\mathrm{T}$ in eq. $(i).$ we get

${2 m g-m g=2 m g(1-\cos \theta)} $

or  ${1-\cos \theta=\frac{1}{2}}$

$ \cos \theta =\frac{1}{2} $

$ \theta =60^{\circ} $