Answer
See the detailed answer below.
Work Step by Step
We call the left upper point 1, the point below it 2, the lower right point 3, and the last point 4 (above 3). See the figure below.
Now we need to find the number of moles of the gas using the ideal gas law and the $-23^\circ \rm C$-point [Point 4].
$$n=\dfrac{P_4V_4}{RT_4}=\dfrac{(150\times 10^3)(100\times 10^{-6})}{(8.31)(-23+273)}$$
$$n=\color{blue}{\bf 0.00722}\;\rm mol$$
The helium gas is a monatomic gas, so $C_V=3R/2$ and $C_P=5R/2$, and hence $\gamma=5/3 $.
Now we also need to find $T_1$, $T_2$.
From 4 to 1, it is an adiabatic process, so
$$T_4V_4^{\gamma-1}=T_1 V_1^{\gamma-1}$$
Hence,
$$T_1=T_4\left[\dfrac{V_4}{V_1}\right]^{\gamma-1}=(-23+273)\left[\dfrac{100}{40}\right]^{\frac{5}{3}-1}=\color{blue}{\bf 460.5}\;\rm K$$
By the same approach, using the process from 2 to 3.
$$T_2=T_3\left[\dfrac{V_3}{V_2}\right]^{\gamma-1}=(-73+273)\left[\dfrac{100}{40}\right]^{\frac{5}{3}-1}=\color{blue}{\bf 368.4}\;\rm K$$
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b) The work output for one cycle is given by
$$W_{out}=(W_s)_{1\rightarrow2}+(W_s)_{2\rightarrow3}+(W_s)_{3\rightarrow4}+(W_s)_{4\rightarrow1}$$
we know that through the isochoric processes, the work done is zero.
$$W_{out}=0+(W_s)_{2\rightarrow3}+0+(W_s)_{4\rightarrow1}$$
And we know that the heat exchange during the adiabatic processes is zero, so the work done by the system is then given by $W_s=-\Delta E_{th}$.
$$W_{out}=-(\Delta E_{th})_{2\rightarrow3} -(\Delta E_{th})_{4\rightarrow1}$$
$$W_{out}=-\frac{3}{2}nR(T_3-T_2) -\frac{3}{2}nR(T_1-T_4)$$
$$W_{out}=-\frac{3}{2}nR [(T_3-T_2) +(T_1-T_4)]$$
$$W_{out}=-\frac{3}{2}nR [T_3-T_2+T_1-T_4]$$
$$W_{out}=-\frac{3}{2}(0.00722)(8.31) [(-73+273)-368.4+460.5-(-23+273)]$$
$$W_{out}=-3.79\;\rm J$$
Thus,
$$W_{in}=-W_{out}=\bf 3.79\;\rm J$$
Hence, the power input for 60 cycles per second is
$$P_{\rm input}=-3.79\times 60=\color{red}{\bf 227.4}\;\rm W$$
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a) The coefficient of performance is given by
$$K=\dfrac{Q_C}{W_{in}}=\dfrac{Q_C}{3.79}$$
Where $Q_C=Q_{3\rightarrow 4}$ since it is the only process at which the gas gained heat. It loses heat in the process of 1 to 2.
$$K= \dfrac{\frac{3}{2}nR(T_4-T_3)}{3.79}$$
$$K= \dfrac{\frac{3}{2}(0.00722)(8.31)(-23-[-73])}{3.79}=\color{red}{\bf 1.19}$$
