Piston $B$ is fixed i.e. it is isochoric process.
If same amount of heat $\Delta Q$ is given to both then
${(\Delta Q)_{{\rm{isobaric}}}} = {(\Delta Q)_{{\rm{isochoric}}}}$==> $\mu \,{C_p}{(\Delta T)_A} = \mu \,{C_v}{(\Delta T)_B}$
==>${(\Delta T)_B} = \frac{{{C_p}}}{{{C_v}}}{(\Delta T)_A} = \gamma {(\Delta T)_A} = 1.4 \times 30 = 42\,K.$


The correct option ($s$) is (are)
$(A)$ $q_{A C}=\Delta U_{B C}$ and $W_{A B}=P_2\left(V_2-V_1\right)$ $(B)$ $W _{ BC }= P _2\left( V _2- V _1\right)$ and $q _{ BC }= H _{ AC }$ $(C)$ $\Delta H _{ CA }<\Delta U _{ CA }$ and $q _{ AC }=\Delta U _{ BC }$ $(D)$ $q_{B C}=\Delta H_{A C}$ and $\Delta H_{C A}>\Delta U_{C A}$