Resistance Multiplier
The way that I would solve this is to divide it into two parts, the photodiode circuit and the resistor divider. 

A photo diode circuit (if R1 and R3 were removed would be:
\$ V_{out} = R_b *(-i_p) \$

A voltage divider from V2 to Vout is:
\$ V_2 = V_{out}*\frac{R_3}{R_1+R_3} \$

rearrange:

\$ V_{out} = V_2*\frac{R_1+R_3}{R_3} \$

Then I just take a shortcut and consider the voltage divider to be a "gain" in the feedback path and substitute

\$ V_{out} = R_b i_p \frac{R_1+R_3}{R_3} =  R_b i_p (1+\frac{R_1}{R_3})\$

You could also use node voltage but I'm too lazy, if you do, the V- terminal would be zero (for an op amp in feedback), then you consider the only path of \$i_p \$ to go through \$ R_p\$ then solve V_2 and Vout and end up with only a Vout equation

Edit:

Soruce transformation, since there is no current going to V- the equation for the node V2 would be this:

\$ V_{2} = R_b -i_p \$

Why? (although I drew the current in the wrong direction, but you get the drift)







simulate this circuit &ndash; Schematic created using CircuitLab

so 
\$ V_{2} = -R_b i_p \$

\$ V_2 = V_{out}*\frac{R_3}{R_1+R_3} \$