From triangle ACP:

$PC = 52 \sin 30^\circ = 26 ~ \text{m}$
$\alpha = 60^\circ - \beta$

From triangle BCP:

$\cos \beta = \dfrac{26}{30}$
$\beta = 30^\circ$

$\alpha = 60^\circ - 30^\circ = 30^\circ$

From triangle PFE:

$\cos \theta = \dfrac{15}{30}$
$\theta = 60^\circ$

$\phi = 180^\circ - \alpha - 2\beta - \theta$

$\phi = 180^\circ - 30^\circ - 2(30^\circ) - 60^\circ$

$\phi = 30^\circ$

Required Area:

$A = A_{\text{sector } BPG} + A_{\text{triangle } BPD} + A_{\text{sector } DPE} + A_{\text{triangle } EPF}$

$A = \dfrac{\pi(30^2)(30^\circ)}{360^\circ} + \dfrac{30^2 \sin 60^\circ}{2} + \dfrac{\pi(30^2)(30^\circ)}{360^\circ} + \dfrac{15(30) \sin 60^\circ}{2}$

$A = 1055.8 ~ \text{m}^2$ *answer*