Once we accept that the mystery reactant is $\ce{P2I4}$, this is the equation we are asked to balance (this might or might not be a fantasy reaction - the question does not teach you much about chemistry):
$$\ce{vP2I4 + w P4 + x H2O -> y PH4I + z H3PO4}$$
This is a problem with 5 unknowns. You can come up with 5 constraints (the number of $\ce{P, I, H, O}$ atoms on either side has to be equal, and the coefficients should be the smallest non-zero integers). So unless any of these conditions are redundant, there should be a single solution.
I will show a simple way of balancing this without using half-reactions (even though this is a redox reaction). Then, I will discuss the complexity of trying to use half-reactions.
Using the atom balance to find the coefficients
Notice that all species are neutral, so we have an automatic charge balance. I will start with an atom present in only two species, oxygen, to eliminate the variable $x$, then proceed to hydrogen, iodine, and finally phosphorus:
$$\ce{vP2I4 + wP4 + 4zH2O -> y PH4I + z H3PO4}\tag{O}$$
$$\ce{vP2I4 + wP4 + 4zH2O -> 5/4z PH4I + z H3PO4}\tag{H}$$
At this stage, I will multiply by 4 to remove the fraction:
$$\ce{4vP2I4 + 4w P4 + 16z H2O -> 5z PH4I + 4z H3PO4}$$
$$\ce{5/4z P2I4 + 4w P4 + 16z H2O -> 5z PH4I + 4z H3PO4}\tag{I}$$
At this stage, I will multiply by 4 to remove the fraction:
$$\ce{5z P2I4 + 16w P4 + 64z H2O -> 20z PH4I + 16z H3PO4}$$
$$\ce{5z P2I4 + 13/2z P4 + 64z H2O -> 20z PH4I + 16z H3PO4}\tag{P}$$
At this stage, I will set $z$ to 2 to get the lowest integers:
$$\ce{10 P2I4 + 13 P4 + 128 H2O -> 40 PH4I + 32 H3PO4}\tag{done}$$
Less formal way of writing
If you solve this by inspection, you would leave out the unknowns and just write the coefficients in the correct ratios:
$$\ce{4 H2O -> H3PO4}\tag{O}$$
$$\ce{16 H2O -> 5 PH4I + 4 H3PO4}\tag{H}$$
$$\ce{5 P2I4 + 64 H2O -> 20 PH4I + 16 H3PO4}\tag{I}$$
$$\ce{10 P2I4 + 13 P4 + 128 H2O -> 40 PH4I + 32 H3PO4}\tag{P}$$
Half-reaction method
The only atoms that have distinct oxidation states are the phosphorus atoms; there are four species, each one with a different oxidation state. The two reactants could either be oxidized to the +5 state (in $\ce{H3PO4}$) or reduced to the -3 state (in $\ce{PH4I}$). Knowing the coefficients of the balanced equation, we can make arbitrary decisions about how to write the half-reactions.
If we want to reduce all of the $\ce{P2I4}$, we get the following scheme:
The half-reactions would be:
$$\ce{10 P2I4 +100 e- + 80 H+ -> 20 PH4I + 20 I- }\tag{red}$$
$$\ce{13 P4 + 20 I- + 128 H2O -> 20 PH4I + 32 H3PO4 + 80 H+ + 100 e-}\tag{ox}$$
If we want to oxidize all of the $\ce{P2I4}$, we get the following scheme:
I will leave it as an exercise to figure the half-reaction for this scenario. Or you can choose anything in between these two schemes, for example:
Now, both half-reactions contain both reactant species.
So you have non-unique half-reactions, but a unique solution to the combined reaction. To get from the half-reactions, you would have to combine them in a way that no ions are "left over". This makes the half-reaction scheme complicated and confusing in this case.
Redundant conditions
This problem is different than being asked to balance a reaction like this:
$$\ce{H2(g) + O2(g) + NaCl(s) -> H2O(l) + NaCl(aq)}$$
Like the original problem, we have four different types of atoms and five coefficients. However, the atom balance for sodium and for chlorine counts as only one constraint because they are redundant. In this case, the equation can be separated into:
$$\ce{H2(g) + O2(g) -> H2O(l) }$$
$$\ce{NaCl(s) -> NaCl(aq) }$$
and there is no unique way to balance the sum of two independent reactions. For the reaction given by the OP, balancing the reaction has a unique solution, but writing down the half-reactions doesn't. This is unusual and probably disturbing to some.
