Thermodynamically, the derivation of $C_{p} = T \left( \frac{\partial S}{\partial T} \right)_{N,P}$

Question

Beginning with the expression $C_{p} = \left( \frac{\partial H}{\partial T}\right)_{N,p}$, I use the Leibniz rule to get: $$ C_{p} = \left( \frac{\partial H}{\partial S}\right)_{N,p} \left( \frac{\partial S}{\partial T}\right)_{N,p} $$

Why is it true that $\left( \frac{\partial H}{\partial S}\right)_{N,p} = T$? I don't see how this is the case; I know that the definition of temperature is $\left( \frac{\partial U}{\partial S}\right)_{N,V} = T$, but I'm not sure how it can possibly relate to this.

EDIT: I meant the derivative of enthalpy $H = U + PV$, not just the energy.

Answer 1

$\text{d}H=T\text{d}S+V\text{d}P$, then $\left(\frac{\partial H}{\partial S}\right)_P=T$, doesn't it?