How to draw the structure and find the oxidation number of Phosphorous in H2P2O7
If you didn't understand what  Poutnik has sain in his comment, this is how it should be done:
You have realized your parent compound is $\ce{H4P2O7}$ (adding two $\ce{H+}$). Correct? If so, concentrate on following hypothetical reaction:
$$\ce{2H3PO4 -&gt; H4P2O7 + H2O} \tag1$$
I believe you know what $\ce{H3PO4}$ is looks like. Now, draw two $\ce{H3PO4}$ side by side. Choose two closest $\ce{OH}$ groups (that would be $\ce{-P-\color{red}{OH} \space \space \color{tequish}{HO}-P -}$). Then, remove one $\ce{\color{red}{-OH}}$ and the $\ce{\color{tequish}{-H}}$ from the second $\ce{\color{tequish}{-OH}}$ you chose to eliminate a $\ce{H2O}$ molecule. Finally combine two remaining molecules together such that final structure forms $\ce{-P\color{red}{-}\color{tequish}{O}-P -}$ bond. This is your $\ce{H4P2O7}$. Eliminate any two hydrogens to get $\ce{H2P2O7^2-}$ structure.
Note: Since above hypothetical equation $(1)$ does not require oxidation or reduction to form $\ce{H4P2O7}$, it is safe to assume that oxidation state of $\ce{P}$ in both $\ce{H3PO4}$ and $\ce{H4P2O7}$ is same.