Why is this BJT transistor saturated?
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.