The correct order of pK$$_a$$ values for the following compounds is:

The four compounds are

a : $$CH_3-CH(OH)-CH_3$$  (isopropyl alcohol, 2° alcohol)
b : $$(CH_3)_3C-OH$$  (tert-butyl alcohol, 3° alcohol)
c : $$C_6H_5OH$$  (phenol)
d : $$CH_3CH_2OH$$  (ethanol, 1° alcohol)

Acid strength is judged by the stability of the conjugate base. A smaller $$pK_a$$ means a stronger acid; a larger $$pK_a$$ means a weaker acid.

Step 1 — Alcohols (a, b, d)
Removal of $$H^+$$ from an alcohol gives an alkoxide ion $$(RO^-)$$.  Alkyl groups are electron-releasing (+I) and therefore destabilise the negatively charged oxygen.  The more alkyl groups attached to the carbon bearing the $$OH$$, the greater the +I effect and the weaker the acid.

Hence for simple aliphatic alcohols:
$$3^\circ \text{ alcohol (b)} \; \lt \; 2^\circ \text{ alcohol (a)} \; \lt \; 1^\circ \text{ alcohol (d)}$$ in acidity, or, in terms of $$pK_a$$ (remember, weaker acid ⇒ larger $$pK_a$$):
$$pK_a(b) \; \gt \; pK_a(a) \; \gt \; pK_a(d)$$

Step 2 — Phenol (c)
For phenol, deprotonation gives the phenoxide ion which is resonance-stabilised:

$$C_6H_5OH \;\rightarrow\; C_6H_5O^- + H^+$$

The negative charge can delocalise over the aromatic ring, greatly stabilising the anion. This resonance stabilisation outweighs the +I effect of the ring, so phenol is far more acidic than normal alcohols. Consequently its $$pK_a$$ is much smaller.

Step 3 — Combining the two trends
Putting phenol together with the alcohols, the overall acidity order is

phenol (c) > ethanol (d) > isopropanol (a) > tert-butanol (b)

Converting to $$pK_a$$ values (largest value for the weakest acid):

$$pK_a(b) \; \gt \; pK_a(a) \; \gt \; pK_a(d) \; \gt \; pK_a(c)$$

This matches Option C:

Option C which is: b > a > d > c