1. We need to count the number of bridging $\mathrm{P-O-P}$ linkages in each compound.

A $\mathrm{P-O-P}$ bond means one oxygen atom is shared between two phosphorus atoms.

2. For $\mathrm{H_4P_2O_7}$ (pyrophosphoric acid)

Its structure is obtained by condensation of two $\mathrm{H_3PO_4}$ molecules:

So the structure contains exactly one bridging oxygen between two phosphorus atoms:

Hence, number of $\mathrm{P-O-P}$ bonds $=1$.

3. For $(\mathrm{HPO_3})_3$

This is metaphosphoric acid in cyclic trimer form. It forms a ring with three phosphorus atoms connected through three bridging oxygens:

closing into a cycle.

Thus, there are $3$ distinct $\mathrm{P-O-P}$ linkages.

Hence, number of $\mathrm{P-O-P}$ bonds $=3$.

4. For $\mathrm{P_4O_{10}}$

$\mathrm{P_4O_{10}}$ has a cage structure derived from $\mathrm{P_4O_6}$. In this structure, the four phosphorus atoms are connected through six bridging oxygen atoms, and each phosphorus also has one terminal $\mathrm{P=O}$ bond.

So the number of bridging $\mathrm{P-O-P}$ linkages is $6$.

Hence, number of $\mathrm{P-O-P}$ bonds $=6$.

5. Therefore, the required numbers are:

This matches option $\boxed{A}$.

6. Comparison with stored correct answer

Stored correct answer: $A$

Our derived answer: $A$

So, the answers agree.