How to measure DC Operating Point of a simple transistor circuit
I assume you use the base as an input.

The base potential is set by the voltage divider \$R_{b1}\$ and \$R_{b2}\$. \$V_b=V_t\cdot\frac{R_{b2}}{R_{b1}+R_{b2}}=15\cdot\frac{56k}{166k}\approx5.1V\$. The base-emitter junction is like a forward-biased diode, dropping about 0.7 volts, therefore the emitter potential is \$V_e=V_b-0.7=4.4V\$.

Now the emitter bias current (which is approximately the same as the collector current) can be calculated as \$I_{e0}=\frac{V_e}{R_e}=\frac{4.4}{5.1k}\approx0.86mA\$

The voltage drop from the collector to the emitter is approximately the following:

\$V_{ce0}\approx V_t-V_e-I_{e0}\cdot R_c=15-4.4-0.86\cdot5.1=15-8.8=6.2 V\$