The solution is the row-selector, trying to maximize the value V, chooses row 1 25% of the time and row 3 75% of the time. The column-selector, trying to minimize V, chooses column 1 half the time and column 3 half the time.
The value of the game (assuming the row selector is trying to maximize) is 5/2.
Your equations miss the point: the right equations to use are of the form (for example) of
$$
V \leq P_1 - P_2 + 3P_3
$$
$$
V \leq -P_1 +5 P_2 + 3P_3
$$
$$
V \leq 4P_1 +3 P_2 + 2P_3
$$
$$
P_1 + P_2 + P_3 = 1
$$
and maximize V subject to all variables $P_i \geq 0$.
For this specific game, you can immediately tell that the second column is dominated by the first, and after eliminating the second column, that the second row is dominated by the first, leaving a 2x2 game with the solution as given.
Although the general way to solve such games is the simplex method, and that actually is not too ugly to do for a 3x3 game matrix, my mantra when given a 3x3 game is:
Form, for the rows, two extra twos consisting of the top row minus the second, then top minus the third. Among those two rows, write the determinant of the (1,2) columns square matrix under the third column, the determinant of the (2,3) square under the first column, and the determinant of the (3,1) column square under the second column.
Then do the same for the columns, subtracting columns 2 and 3 from the first.
$$
\begin{array}{c c c | c c | c}
1 & 2 & 4 & -1 & -3 & -6 \\
-1 & 5 & 3 & -6 & -4 & +1 \\
3 & 3 & 2 & 0 & 1 & -14 \\
\hline
2 & -3 & 1 & & & \\
-2 & -1 & 2 & & & \\
\hline
-5 & -6 & -8 & & &
\end{array}
$$
If the answers for the rows are all the same sign, and the answers for the columns are all the same sign, then the game solution requires use of all three rows and all three columns, and you read off the strategies as the ratios of the numbers you just found. But in this case, that does not work because the column numbers are -6, +1, and -14. That means that the correct solution will use fewer than three rows.
Try eliminating one row and solving the resulting 2x3 game: First you might try eliminating row 1. But the solution to that game (of value 11/5) has the column chooser selecting column 3 80% of the time and column 1 20%; clearly the row chooser can do better than 11/5 by picking row 1.
When we eliminate row 2, we get the right solution, which is the solution presented above.
It is easy to check that neither player can improve his result by choosing one of the eliminated choices.
In some rare cases, you have a situation where there are a family of correct solutions, but each member of the family requires a mix of all 3 rows (or columns) while the strategy for the other player is always the same ratio of only two of the columns (or rows). This makes the solution less obvious, although the simplex method always works. I may present such a game as a problem on this site at a later date.
Your observations are correct, though they apply quite specifically to your problem. It is not uncommon for the method of Lagrange multipliers to yield equations that either you already knew, or are useless (like $0 = 0$).
What is true in general is that you never have to use the method of Lagrange multipliers. It's always possible (perhaps not algebraically, but definitely numerically) to use the constraint to eliminate one of the variables, but this method may be disadvantageous for a couple reasons (it may complicate the calculations, for one). For your problem, in many cases we could use the constraint $f(\alpha, z) = c$ and solve for $z$ as a function of $c$ and $\alpha$ and then set the derivative of $z$ with respect to $\alpha$ to zero like in a normal one-variable extremization problem. These will lead to the exact same equations you have already deduced.
The moral of the story? There is no hands-down most efficient way for solving many extremization problems; it will depend on the nature of the problem.
Best Answer
Let $f(p)=\sum_{k=0}^4 k^2 p_k$.
Let $g_k(p)=-p_k$ and $\overline{g}_k(p)=p_k-1,k=0,1,\dots,4$. (Bars in this answer aren't conjugates, it's just notation.)
Let $h_0(p)=\sum_{k=0}^4 p_k-1,h_1(p)=\sum_{k=0}^4 k p_k - A$.
In this notation, the KKT stationarity condition is
$$\pm \nabla f(p)=\sum_{k=0}^4 \left ( \mu_k \nabla g_k(p) + \overline{\mu}_k \nabla \overline{g}_k(p) \right ) + \sum_{k=0}^1 \lambda_k \nabla h_k(p)$$
with the sign depending on whether you are maximizing (+) or minimizing (-).
You also have the constraints of course (called "primal feasibility" conditions), which in this notation read:
$$h_k=0,k=0,1 \\ g_k \leq 0,k=0,1,\dots,4 \\ \overline{g}_k \leq 0,k=0,1,\dots,4.$$
Finally you have the other two KKT conditions. First there's dual feasibility:
$$\mu_k \geq 0,k=0,1,\dots,4\\ \overline{\mu}_k \geq 0,k=0,1,\dots,4.$$
Second there's complementary slackness:
$$\mu_k g_k=0,k=0,1,\dots,4 \\ \overline{\mu}_k \overline{g}_k=0,k=0,1,\dots,4.$$
In total you have 17 equality constraints in 17 unknowns, so the equation counting checks out.