Why KCL does not apply in case of this bjt circuit?
Kirchhoff's voltage law applies in setting up the DC operating point:

$$\begin{align*}
9\:\textrm{V}-I_E\cdot R_C-I_B\cdot R_B-V_{BE}&amp;=0\:\textrm{V} \\\\
9\:\textrm{V}-\left(\beta+1\right)\cdot I_B\cdot R_C-I_B\cdot R_B-V_{BE}&amp;=0\:\textrm{V} \\\\
9\:\textrm{V}-V_{BE}&amp;=\left(\beta+1\right)\cdot I_B\cdot R_C+I_B\cdot R_B \\\\
9\:\textrm{V}-V_{BE}&amp;=I_B\cdot\left[\left(\beta+1\right)\cdot R_C+R_B\right] \\\\
I_B&amp;=\frac{9\:\textrm{V}-V_{BE}}{\left(\beta+1\right)\cdot R_C+R_B}
\end{align*}$$

Estimating \$V_{BE}\approx 700\:\textrm{V}\$, this means \$I_B\approx 11.5\:\mu\textrm{A}\$ and \$I_C\approx 2.3\:\textrm{mA}\$. And therefore \$V_C= 9\:\textrm{V}-2.3\:\textrm{mA}\cdot 2.7\:\textrm{k}\Omega\approx 2.8\:\textrm{V}\$.

I get the following pair of simultaneous equations using Kirchhoff's current law, instead, for the DC operating point:

$$\begin{align*}
\frac{V_C}{R_C}+\frac{V_C}{R_B}+\beta\cdot I_B&amp;=\frac{9\:\textrm{V}}{R_C}+\frac{V_{BE}}{R_B}\\\\
\frac{V_{BE}}{R_B}+I_B&amp;=\frac{V_C}{R_B}
\end{align*}$$

Using \$I_B\$ and \$V_C\$ as unknowns, the above two equations simultaneously solve out to the exact same resulting equation for \$I_B\$ shown above. And the rest of the details fall out in the usual way, as well.

It all just works.