We have $\left|\frac{z-p}{\bar{p} \cdot z-1} \right|^2 = \frac{(z-p) \cdot \overline{(z-p)}}{(\bar{p} \cdot z-1) \cdot \overline{(\bar{p} \cdot z-1)}} \\ = \frac{z \cdot \bar{z}-p \cdot \bar{z} - \bar{p} \cdot z + p \cdot \bar{p}}{z \cdot \bar{z} \cdot \bar{p} \cdot p- \bar{p} \cdot z - p \cdot \bar{z}+1}$
Hence $\left|\frac{z-p}{\bar{p} \cdot z-1} \right|^2 < 1 \\ \Leftrightarrow z \cdot \bar{z}-p \cdot \bar{z} - \bar{p} \cdot z + p \cdot \bar{p} < z \cdot \bar{z} \cdot \bar{p} \cdot p- \bar{p} \cdot z - p \cdot \bar{z}+1 \\ \Leftrightarrow |z|^2 \cdot (1-|p|^2) < (1-|p|^2) \\ \Leftrightarrow |z|<1$
So the first option is true. Similar proof shows that the second one is true (remark: $1-|p|^2 <0$ for $|p|>1$). The third one can't be true since pre-images of closed subsets are closed ($f$ is continuous). The last one is clearly false.
Remark The given mapping is a Möbius transformation. This class of functions has some nice properties...