In a molecule of pyrophosphoric acid, the number of P - O - H, P = O and P - O - P bonds/moiety(ies) respectively are:

First, we recall that the molecular formula of pyrophosphoric acid is $$\mathrm{H_4P_2O_7}$$. This molecule is produced by the condensation of two molecules of orthophosphoric acid $$\bigl(\mathrm{H_3PO_4}\bigr)$$ with the elimination of one molecule of water:

$$2\,\mathrm{H_3PO_4}\;\longrightarrow\;\mathrm{H_4P_2O_7}+ \mathrm{H_2O}$$

This dehydration joins the two $$\mathrm{PO_4^{3-}}$$ tetrahedra through one common oxygen atom. Hence, in the final structure each phosphorus atom remains tetra-coordinated, exactly as in orthophosphoric acid, but now one of the single bonds on each phosphorus is to the same bridging oxygen. We can sketch the connectivity as

$$\mathrm{HO\!-\!P(=O)(OH)\!-\!O\!-\!P(=O)(OH)\!-\!OH}$$

Now we simply count the different kinds of linkages present in this expanded formula.

Counting the $$\mathbf{P-O-H}$$ (hydroxyl) bonds: We see four separate “OH” groups: the two on the left phosphorus and the two on the right phosphorus. So,

$$\text{Number of }P-O-H \text{ bonds}=4$$

Counting the $$\mathbf{P = O}$$ (doubly bonded oxygen) bonds: Each phosphorus is still double-bonded to one oxygen atom, just as in orthophosphoric acid. Since there are two phosphorus atoms, we get

$$\text{Number of }P = O \text{ bonds}=2$$

Counting the $$\mathbf{P-O-P}$$ (bridging oxygen) linkage: There is exactly one oxygen that is shared between the two phosphorus atoms, giving exactly one $$P-O-P$$ bridge.

$$\text{Number of }P-O-P \text{ bonds}=1$$

Collecting these results, the set of numbers is $$4,\,2,\,1$$ in the order asked. This matches Option D.

Hence, the correct answer is Option D.