$AP = {2 \over 3}AD = {8 \over 3};\,\,PD = {4 \over 3};\,\,$ Let $PB=x$ $\tan {60^ \circ } = {{8/3} \over x}$ or $x = {8 \over {3\sqrt 3 }}$ Area of $\Delta ABD$ $= {1 \over 2} \times 4 \times {8 \over {3\sqrt 3 }} = {{16} \over {3\sqrt 3 }}$ $\therefore$ Area of $\Delta ABC$ $= 2 \times {{16} \over {3\sqrt 3 }} = {{32} \over {3\sqrt 3 }}$ $\left[ \, \right.$ As median of a $\Delta$ divides it into two $\Delta 's$ of equal area. $\left. \, \right]$