1. Use the spontaneity criteria from thermodynamics

For an irreversible (spontaneous) process:

• At constant $E$ and $V$, entropy of the system satisfies $\left(dS\right)_{E,V} > 0$
• At constant $T$ and $P$, Gibbs free energy satisfies $\left(dG\right)_{T,P} < 0$

These are standard equilibrium criteria.

2. Why these conditions apply

Since the question states:

• process is irreversible,
• takes place at constant $T$ and $P$,
• only pressure-volume work is done,

for a spontaneous change at constant temperature and pressure, $dG < 0$ with equality only at equilibrium.

Also, the corresponding entropy criterion is that for an isolated description at constant energy and volume, $\left(dS\right)_{E,V} > 0$ for a spontaneous irreversible process.

3. Match with the options

• Option A: $\left(dS\right)_{E,V} > 0, \quad \left(dG\right)_{T,P} < 0$ This is correct.

• Option B: $\left(dS\right)_{E,V} = 0, \quad \left(dG\right)_{T,P} = 0$ This corresponds to equilibrium, not an irreversible spontaneous process.

• Option C: $\left(dS\right)_{E,V} = 0, \quad \left(dG\right)_{T,P} > 0$ Not spontaneous.

• Option D: $\left(dS\right)_{E,V} < 0, \quad \left(dG\right)_{T,P} < 0$ First condition is wrong for spontaneity.

4. Final answer

Therefore, the correct option is: $\boxed{\text{A}}$