At equilibrium, we have  $[{N_2}] = 0.79\,(1 - \alpha );$

$[{O_2}] = 0.21\,(1 - \alpha );\,[NO] = 2\alpha $

Total number of moles

$ = 0.79(1 - \alpha ) + 0.21(1 - \alpha )9 + 2\alpha  = 1 + \alpha $

${P_{{N_2}}} = \frac{{0.79(1 - \alpha )}}{{1 + \alpha }} \times 1;$

${P_{{O_2}}} = \frac{{0.21(1 - \alpha )}}{{1 + \alpha }} \times 1;\,{P_{NO}} = \frac{{2\alpha }}{{1 + \alpha }} \times 1$

${K_P} = \frac{{P_{NO}^2}}{{{P_{{N_2}}}.{P_{{O_2}}}}}$

$1.1 \times {10^{ - 3}} = \frac{{4{\alpha ^2}}}{{0.79 \times 0.21{{(1 - \alpha )}^2}}}$

or $\alpha  = 0.0067$

$ \Rightarrow $ vol $\%$ of $NO = 2\alpha  \times 100$

                        $ = 2 \times 0.0067 \times 100 = 1.33\,\% $