Solution:
Extend $PM$ to a point $C$ such that $PC \perp OC$. Since $\angle PMO = 120^\circ$, $\angle CMO = 60^\circ$ and $\angle COM = 30^\circ$. Let $PB = x$ and $MC = a$. Then $CO = a \sqrt{3}$ and $OM = 2a$. Moreover, $\triangle PMB$ and $\triangle PCO$ are similar triangles. Thus, we have
$$\frac{2}{2+a} = \frac{x}{x+2}$$
so $x = \frac{4}{a}$.

Furthermore, by the Cosine Law on side $PO$ of $\triangle PMO$, we have
$$(x+2)^2 = 4 + 4a^2 + 2a$$
Plugging in $x = \frac{4}{a}$ and expanding, we have
$$\frac{16}{a^2} + \frac{16}{a} + 4 = 4 + 4a^2 + 4a$$
and so $4 + 4a = a^4 + a^3$. Hence $a^3 = 4$ and $a = \sqrt[3]{4}$. Thus, $x = \frac{4}{\sqrt[3]{4}}$.

It follows that the product of the lengths of the sides of the triangle is
$$(2a)(2)(x+2) = (2\sqrt[3]{4})(2)(2\sqrt[3]{2} + 2) = 16 + 8\sqrt[3]{4}$$
so $a + b + c = 16 + 8 + 4 = 28$.