Statement (A): Penetration power of s , p , d and f sub shells are s>p>d>f.

Statement (B): ${\mathrm{IE}}_1 \mathrm{of} \mathrm{N}>{\mathrm{IE}}_1 \mathrm{of} \mathrm{O} \mathrm{but} {\mathrm{IE}}_2 \mathrm{of} \mathrm{O}>{\mathrm{IE}}_2 \mathrm{of} \mathrm{N}.$

Statement (C): ${\mathrm{IE}}_3>{\mathrm{IE}}_2>{\mathrm{IE}}_1$

Statement-A: The IE depends on the type of the sub shell in which the electron is present. As s-electron penetrates more nearer to nucleus, so its more tightly held than p electrons and similarly d-electrons are more tightly held then f-electron. Therefore the statement is correct.

Statement -B: $\mathrm{N}=7; 1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^3$ here nitrogen has half-filled stable configuration while for $\mathrm{O}=8; 1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^4$ Oxygen has incompletely filled unstable 2p orbital but for the second IE, ${\mathrm{N}}^{+}= 1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^2$ has incompletely filled unstable 2p orbital whereas for ${\mathrm{O}}^{+}=1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^3$ oxygen has half filled stable configuration. Therefore the statement-2 is correct 

Statement (C): ${\mathrm{IE}}_3>{\mathrm{IE}}_2>{\mathrm{IE}}_1$ is correct because the successive IE of various of electrons is directly proportional to the charge on the residual charge.