已知\(P\)是椭圆\(C:\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1\left({a>b>0}\right)\)上的点，\(F_{1}\)，\(F_{2}\)分别是\(C\)的左，右焦点，\(O\)是坐标原点，若\(\left|{\overrightarrow {OP}+\overrightarrow {OF}_{2}}\right|=2\left|{\overrightarrow {OF}_{1}}\right|\)且\(∠F_{1}PF_{2}=60°\)，则椭圆的离心率为\((\quad)\)