The oxidation states of $$P$$ in H$$_4$$P$$_2$$O$$_7$$, H$$_4$$P$$_2$$O$$_5$$ and H$$_4$$P$$_2$$O$$_6$$, respectively, are:

For any electrically neutral molecule, the sum of the oxidation numbers of all the atoms present is zero. We shall use the usual convention that hydrogen has an oxidation number of $$+1$$ and oxygen has an oxidation number of $$-2$$.

First we consider $$\mathrm{H_4P_2O_7}.$$ Let the oxidation number of each phosphorus atom be $$x.$$ Writing the algebraic sum, we have

$$4(+1) + 2(x) + 7(-2) = 0.$$

This simplifies to $$4 + 2x - 14 = 0,$$ so $$2x - 10 = 0,$$ giving $$2x = 10$$ and finally $$x = +5.$$ Thus, in $$\mathrm{H_4P_2O_7},$$ each phosphorus is in the $$+5$$ oxidation state.

Next we take $$\mathrm{H_4P_2O_5}.$$ Again letting the oxidation number of phosphorus be $$x,$$ we write

$$4(+1) + 2(x) + 5(-2) = 0.$$

Therefore $$4 + 2x - 10 = 0,$$ which gives $$2x - 6 = 0,$$ so $$2x = 6$$ and hence $$x = +3.$$ Consequently, phosphorus is in the $$+3$$ state in $$\mathrm{H_4P_2O_5}.$$

Finally we analyse $$\mathrm{H_4P_2O_6}.$$ Setting the oxidation number of phosphorus equal to $$x,$$ we write

$$4(+1) + 2(x) + 6(-2) = 0.$$

Simplifying, $$4 + 2x - 12 = 0,$$ leading to $$2x - 8 = 0,$$ so $$2x = 8$$ and hence $$x = +4.$$ Thus, in $$\mathrm{H_4P_2O_6},$$ phosphorus has the oxidation state $$+4.$$

Collecting our results, the oxidation states of phosphorus in $$\mathrm{H_4P_2O_7},\; \mathrm{H_4P_2O_5}$$ and $$\mathrm{H_4P_2O_6}$$ are $$+5,\; +3$$ and $$+4$$ respectively.

Hence, the correct answer is Option C.