Answer: A

Solution:

Let $a = BC,$ $b = CA,$ $c = AB.$ Since $E$ is the midpoint of $AC,$ $AE = \tfrac{b}{2}.$ The centroid gives $AG = \tfrac23 m_a$ and $GE = \tfrac13 m_b,$ where $m_a, m_b$ are the medians from $A$ and $B.$

Equilateral $\triangle AGE$ means $AG = GE = AE.$ From $\tfrac23 m_a = \tfrac{b}{2}$ we get $m_a = \tfrac34 b,$ which with $m_a^2 = \tfrac{2b^2 + 2c^2 - a^2}{4}$ gives $2c^2 - a^2 = \tfrac{b^2}{4}.$ From $\tfrac13 m_b = \tfrac{b}{2}$ we get $m_b = \tfrac32 b,$ giving $a^2 + c^2 = 5b^2.$

Solving, $c^2 = \tfrac{7b^2}{4}$ and $a^2 = \tfrac{13b^2}{4}.$ Taking $b = 2$ gives $a^2 = 13,$ $c^2 = 7,$ so $$\begin{aligned} \cos C &= \dfrac{a^2 + b^2 - c^2}{2ab} \\ &= \dfrac{13 + 4 - 7}{2 \cdot \sqrt{13} \cdot 2} \\ &= \dfrac{5\sqrt{13}}{26}. \end{aligned}$$

Then $m + n + p = 5 + 26 + 13 = 44.$

Thus, the correct answer is A.