The given figure shows the P-V diagram for a Carnot cycle. In this diagram,

 

For an adiabatic process, ${\mathrm{PV}}^{\mathrm{\gamma }}=$ constant. Differentiating w.r.t. V, we get 

     $\frac{\mathrm{dP}}{\mathrm{dV}}{\mathrm{V}}^{\mathrm{\gamma }}+{\mathrm{P\gamma V}}^{\mathrm{\gamma }-1}=0$

or   $\frac{\mathrm{dP}}{\mathrm{dV}}=-\frac{\mathrm{\gamma P}}{\mathrm{V}}$

For isothermal process, PV = constant

Hence, 

          $\frac{\mathrm{dP}}{\mathrm{dV}}=-\frac{\mathrm{P}}{\mathrm{V}}$

Now, dP/dV is the slope of the P-V graph. Thus, the slope of the (P-V) graph for an adiabatic process is $\mathrm{\gamma }$ times that for an isothermal process. Hence, curves BC and DA both represent adiabatic process and curves AB and CD both represent isothermal process. Thus, the correct choices are (a) and (c).