In graphite and diamond, the percentage of p characters of the hybrid orbitals in hybridization are respectively:

We first recall the general formula that relates the designation of a hybrid orbital to the percentage of p-character present in it. If an orbital is described as $$\text{sp}^n,$$ this means that one $$s$$ orbital and $$n$$ identical $$p$$ orbitals are mixed. The total number of atomic orbitals mixed is therefore $$1+n.$$ Hence:

Percentage of $$s$$-character $$=\dfrac{1}{1+n}\times 100$$

Percentage of $$p$$-character $$=\dfrac{n}{1+n}\times 100$$

Now we analyse the two carbon allotropes one by one.

In graphite, every carbon atom is trigonal-planar and uses $$\text{sp}^2$$ hybridisation. Here we have $$n=2.$$ Substituting $$n=2$$ in the above expression for the $$p$$-character, we get

$$\text{Percentage of }p\text{ in graphite}= \dfrac{2}{1+2}\times 100=\dfrac{2}{3}\times 100=66.67\%.$$

Rounding to the nearest whole number, this is $$67\%.$$

In diamond, every carbon atom is tetrahedral and uses $$\text{sp}^3$$ hybridisation. Now $$n=3.$$ Substituting $$n=3$$ into the same formula, we have

$$\text{Percentage of }p\text{ in diamond}= \dfrac{3}{1+3}\times 100=\dfrac{3}{4}\times 100=75\%.$$

So the required percentages of $$p$$-character are $$67\%$$ for graphite and $$75\%$$ for diamond.

Hence, the correct answer is Option B.