Unless I miss something, there seem to be many possibilities. Trivially, all payoffs could be the same, in which case every strategy is optimal. Another e.g. let $n=m$ and the payoff matrix be diagonal with entries $\dfrac{V}{p_i}$ for non-zero $V, p_i$. I am sure you can generate many more.
Given a payoff matrix of size n, the gain of a mixed strategy $(p_1, p_2, ... p_n)$ can be expressed as a polynomial (quadratic) function of $(p_1, p_2, ... p_{n-1})$ in the polytope defined by $0\le p_i\le 1$ and $\sum_{1\le i<n}p_i\le 1$.
Finding the optimal strategies is related to find all local maximum of the function : in the general case, you need to evalute the gradient and the eigenvalues of the Hessian matrix (you can read more at https://en.wikipedia.org/wiki/Hessian_matrix#Critical_points)
If you find one maximum inside the polytope, then you have your mixed strategy (there can by only one as the function is quadratic).
If it fails (no maximum inside the polytope) then you need to find maximum on the borders (each part of the border is defined by $p_i=0$ for some $1\le i\le n$) by applying the same method with one less variable. Each border can have its own set of local maximum.
There is always the possibility to have degenerate solution (with eigenvalues = 0) that can imply that the set of optimal strategies is not a set of isolated points (but something larger, like an hyperplan $ p_1=p_2$).
In your example :
$$g(p_1,p_2)=2p_1p_2+4p_1p_3+6p_2p_3 $$ by replacing $p_3=1-p_1-p_2$
$$g(p_1,p_2)=-4p_1^2-8p_1p_2-6p_2^2+4p_1+6p_2$$
the gradient is (first derivatives) $(-8p_1-8p_2+4, -8p_1-12p_2+6)$ that is zero iff $(p_1=0, p_2=\frac{1}{2})$.
The Hessian matrix is a constant matrix (the second derivatives of a quadratic function) and here it's
$$\left[\begin{matrix}
-8 & -8 \\ -8 & -12\\
\end{matrix}\right]$$
The eigenvalues are all negative (as the determinant is positive (32) and the trace is negative (-20)), so it's a global maximum at $(0,\frac{1}{2},\frac{1}{2})$. No need to look at the borders (which can be quite complex).
So there is only one stable and optimal strategy for this game.
Best Answer
The basic assumptions in a game theory are:
You can solve this problem by Reducing by Dominance.
Since the entries in Row 2 are greater (or equal to) the corresponding ones in Row 1 (it implies Row 2 dominates Row 1, i.e. Player 2 is better off with strategy 2), we can eliminate Row 1 to get: $$\begin{array}{cc|cc} &&Player \ 1 \\ &&A&B \\ \hline Player \ 2 & 2 & 8 & 10 \end{array}$$ Since the entry in Column A is less than (or equal to) the corresponding one in Column B (it implies Column A dominates Column B, i.e. the Player 1 is better off with strategy A), we can eliminate Column B to get: $$\begin{array}{cc|c} &&Player \ 1 \\ &&A \\ \hline Player \ 2 & 2 & 8 \end{array}$$ So, the Player 2 has an advantage of $8$ units over Player 1.