You write that the peak base current, with the signal source connected is given by
$$i_B = \frac{5V - 0.62318V}{5.5M\Omega} = 795.78nA $$
But this isn't true (which should be obvious as it's less than the bias current!). What's true is
$$i_{R2} = \frac{5V - 0.62318V}{5.5M\Omega} = 795.78nA \ne i_B$$
The resistor current and base current are not equal. According to KCL at the base node:
$$i_B = i_{R2} + i_S $$
where \$i_S\$ is the current out of the signal voltage source. But you don't know what this current is.
In fact, the base current depends exponentially on the base-emitter voltage. We can estimate the change in base current as follows
$$\frac{i_{B2}}{i_{B1}} = \frac{e^{\frac{v_{BE2}}{V_T}}}{e^{\frac{v_{BE1}}{V_T}}} = e^{\frac{v_{BE2}- v_{BE1}}{V_T}} = e^{\frac{0.62318V - 0.6V}{25mV}} \approx 2.53$$
Thus, the peak base current should be larger than the DC base current by a factor of about 2.53 or
$$i_{B_{peak}} = 2.53 \cdot 800nA = 2.02\mu A$$
This gives a collector current of
$$i_{C_{peak}} = 2.53 \cdot 233\mu A = 589\mu A$$
If this were the actual collector current, the collector voltage would be
$$5V - 589\mu A \cdot 10k\Omega = -0.895V $$
So, yes, the transistor will saturate first.
Connecting the base to the collector makes it act as a diode, and will drop ~0.7V-1V
Also, for more information about this topic please check out this answer
Best Answer
The collector junction of Qp and emitter junction of Qn - forward biased.
The collector junction of Qn and emitter junction of Qp - reverse biased.
now the base current of Qn is, $$I_{Bn} = \frac{V_1 - V_{BEn} - V_{CBp}}{R1+R5+(\beta+1)R4}$$ then, $$I_{Cn} = \mathrm{min}(\beta I_{Bn}, I_{Csat})$$ where $$I_{Csat} = \frac{V_1-V_{CEn}}{R_4+R_3}$$
The voltages you can calculate as.
$$V_{En} = R4\times(I_{Bn}+I_{Cn})$$ $$V_{Cn} = V_1 - R3\times(I_{Cn})$$ $$\text{so on ...}$$