In the first case,

${\mathrm{n}}_{\mathrm{A}\mathrm{r}}=1,{\mathrm{n}}_{\mathrm{C}{\mathrm{O}}_2}=2,{\mathrm{n}}_{{\mathrm{O}}_2}=3,{\mathrm{n}}_{{\mathrm{N}}_{\mathrm{A}}}=4$

So total number of moles $=1+2+3+4=10$

If P in the total pressure, ${\mathrm{P}}_{{\mathrm{O}}_2}=\frac{3\mathrm{P}}{10}$

In the second case,

${\mathrm{n}}_{{\mathrm{O}}_2}=2,{\mathrm{n}}_{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}=9$

As P remains the same, ${\mathrm{P}}_{{\mathrm{O}}_2}=\frac{2\mathrm{P}}{9}$

So decrease in partial pressure of ${\mathrm{O}}_2$

$=\frac{3\mathrm{P}}{10}-\frac{2\mathrm{P}}{9}=\frac{7\mathrm{P}}{90}$

So % decrease in partial pressure of ${\mathrm{O}}_2$

$=\frac{\frac{7\mathrm{P}}{90}}{\frac{3\mathrm{P}}{10}}\times 100=\frac{700}{27}=26\mathrm{\%}$