BJT input connected to another BJT's output and loading effects
Assuming that NPN is still in active region, both \$I_E\$ and \$I_C\$ will be determined by \$I_B\$. 

It is clear that the voltage divider is not affected by addition of PNP, therefore \$V_B\$ doesn't change.

The only assumption you must do here is that \$I_B\$ doesn't change.

Under this assumption:


\$I_E\$ doesn't change
\$I_C\$ doesn't change 


What changes here? Well, the current drawn from PNP's base must flow through NPN. This current flows also through \$R_4\$ increasing the voltage drop on this resistor. This means that \$V_E\$ rises, which reduces \$V_{BE}\$. The reduction in \$V_{BE}\$ tends to make the current through NPN smaller. This effect is called Negative Feedback - the increase in current leads to decrease in current.

The steady state of the mechanism described in the previous paragraph is that the current through NPN will stay the same, which implies the same \$V_E\$. However, since the current drawn from PNP's base is subtracted from  the current trough \$R_3\$, the voltage drop across this resistor will become smaller, which leads to higher \$V_C\$.

Note that \$I_B\$ and \$I_C\$ are independent of \$V_{CB}\$ in active region of operation (neglecting the Early effect), therefore our initial assumption of the same \$I_B\$ was justified.

In summary: 

Negative feedback mechanism leads to constant current through NPN with slight increase in collector's voltage.

Note:

The above discussion, while completely correct in the framework of simple BJT model, has its limitations and errors as compared to real devices (just like any other model). For example, taking into account Early effect leads to an increase in collector's current (this effect is more pronounce in transistors having smaller dimensions). If the transistors have totally different electric characteristics (for example: the base current from PNP is comparable to collector's current of NPN) the answer may change too.

Hope this helps.