Step 1 - State the type of deviation.
A mixture of phenol and aniline shows NEGATIVE DEVIATION from Raoult's law.

Step 2 - Give the reason.
Phenol has an acidic $-OH$ group (a good hydrogen-bond DONOR) and aniline has a basic
$-NH_2$ group (a good hydrogen-bond ACCEPTOR). When the two are mixed, STRONG
INTERMOLECULAR HYDROGEN BONDS are formed between the unlike molecules
(phenol $O-H \cdots N$ aniline).
These new A-B interactions are STRONGER than the A-A interactions in pure phenol and the
B-B interactions in pure aniline.

Step 3 - Consequence for vapour pressure.
Because the molecules are held more tightly, their escaping tendency FALLS. Therefore
$$p_A < p_A^0 x_A \qquad \text{and} \qquad p_B < p_B^0 x_B$$
$$p_{total} < p_A^0 x_A + p_B^0 x_B$$
The observed vapour pressure is LESS than that predicted by Raoult's law - which is
exactly the definition of negative deviation.
Also $\Delta_{mix}H < 0$ (heat is evolved) and $\Delta_{mix}V < 0$ (volume contracts).

Step 4 - Effect on boiling point.
Boiling point is the temperature at which the vapour pressure equals the atmospheric
pressure. Since the vapour pressure of the mixture is LOWERED, a HIGHER temperature is
needed to reach atmospheric pressure.
Hence the BOILING POINT INCREASES (it is higher than that expected from Raoult's law,
and such mixtures form maximum boiling azeotropes).