Recall that a dualizing sheaf is a sheaf $\omega$ which satisfies $H^1(C, F)^\vee = Hom_C(F, \omega)$ for every coherent $F$.

A thing to start with is duality on $P:=\mathbb{P}^n$:
$$ H^1(C, F)^\vee = H^1(P, F)^\vee = Ext^{n-1}_P(F, \omega_P). $$

We now need a natural isomorphism between $Ext^i_P(F, \omega_P)$ and $Hom_C(F, \omega)$ for some $\omega$. A good thing to do to compare $Hom$'s and $Ext$'s on $P$ and $C$ is to find a spectral sequence comparing the two. Let us look at the functors
$$ Coh(P) \to Coh(C) \to Ab $$
where the first functor is $\mathscr{H}om_P(\mathscr{O}_C, -)$ and the second is $Hom_C(F, -)$, their composition $Hom_P(F, -)$. The spectral sequence of this composition is
$$ E^{ij}_2 = Ext^i_C(F, Ext^j_P(\mathscr{O}_X, G)) \Rightarrow Ext^{i+j}_P(F, G). $$
For $i+j=n-1$ and $G=\omega_P$ we have our group $Ext^{n-1}_P(F, \omega)$ on the right hand side, and we have $E^{0,n-1}_2 = Hom_C(F, \mathscr{E}xt^{n-1}_P(\mathscr{O}_X, \omega_P)$, which suggests we should take  $\omega = \mathscr{E}xt^{n-1}_P(\mathscr{O}_X, \omega_P)$. 

To prove that this works, we need to show that $E^{ij}_2 = 0$ for $j<n-1$.