The correct decreasing order of oxidation state of the underlined atom in each molecule is

Let us calculate the oxidation state of the underlined atom in each molecule.

For option (1):
Oxidation state of P in $\text{P}_4\text{O}_6$ is $+3$.
Oxidation state of Cl in $\text{Cl}_2\text{O}_7$ is $+7$.
Oxidation state of Al in $\text{AlH}_3$ is $+3$.
Order: $+3 < +7 > +3$ (Incorrect)

For option (2):
Oxidation state of P in $\text{P}_4\text{O}_{10}$ is $+5$.
Oxidation state of S in $\text{SO}_3$ is $+6$.
Oxidation state of O in $\text{H}_2\text{O}$ is $-2$.
Order: $+5 < +6 > -2$ (Incorrect)

For option (3):
Oxidation state of N in $\text{N}_2\text{O}_5$ is $+5$.
Oxidation state of Al in $\text{Al}_2\text{O}_3$ is $+3$.
Oxidation state of S in $\text{H}_2\text{S}$ is $-2$.
Order: $+5 > +3 > -2$ (Correct)

For option (4):
Oxidation state of Pb in $\text{PbO}_2$ is $+4$.
Oxidation state of N in $\text{N}_2\text{O}_3$ is $+3$.
Oxidation state of S in $\text{SO}_3$ is $+6$.
Order: $+4 > +3 < +6$ (Incorrect)