Step 1 - List the ions present.
$CuCl_2 \rightarrow Cu^{2+} + 2Cl^-$ and $H_2O \rightleftharpoons H^+ + OH^-$
Cations available at the cathode: $Cu^{2+}$, $H^+$
Anions available at the anode: $Cl^-$, $OH^-$ (from water)

Step 2 - Decide the cathode reaction.
Reduction occurs at the cathode. The species with the HIGHER reduction potential is
discharged first.
$$Cu^{2+} + 2e^- \rightarrow Cu(s), \qquad E^\circ = +0.34\ V$$
$$2H_2O + 2e^- \rightarrow H_2 + 2OH^-, \qquad E^\circ = -0.83\ V$$
Copper has the higher value, so Cu is DEPOSITED AT THE CATHODE (not at the anode,
and no $H_2$ is released). This rules out options (A) and (B).

Step 3 - Decide the anode reaction.
Oxidation occurs at the anode. Platinum is inert, so it does not dissolve.
$$2Cl^- \rightarrow Cl_2 + 2e^- \qquad (E^\circ_{ox} = -1.36\ V)$$
$$2H_2O \rightarrow O_2 + 4H^+ + 4e^- \qquad (E^\circ_{ox} = -1.23\ V)$$
On the basis of potential alone $O_2$ should be liberated, but the liberation of oxygen
on a platinum surface requires a large extra voltage called the OVER-POTENTIAL
(over-voltage). Because of this over-potential, $Cl^-$ is oxidised in preference and
$Cl_2$ gas is evolved at the anode.

Step 4 - Conclusion.
Cathode: $Cu$ deposited. Anode: $Cl_2$ gas released.

Answer: (D) $Cl_2$ gas will be released at anode