a) Among any 8 consecutive integers there exists one of the form $8j + 4$. This number has the property ($p$) because the factor 2 from its prime factorization has the exponent 2. Therefore there can be at most 7 consecutive positive integers that do not have the property ($p$). Since none of the numbers 29, 30, 31, 32, 33, 34, 35 has the property ($p$), the largest $k$ is $k = 7$.

b) Since the numbers $98 = 2 \cdot 7^2$, $99 = 3^2 \cdot 11$, and $100 = 2^2 \cdot 5^2$ have the property ($p$), so will the numbers $98 + (7 \cdot 3 \cdot 2)^3 \cdot k$, $99 + (7 \cdot 3 \cdot 2)^3 \cdot k$, and $100 + (7 \cdot 3 \cdot 2)^3 \cdot k$.

Alternative solution.

By the Chinese remainder theorem, there exist an infinite number of solutions for the system of simultaneous congruences: $n \equiv 4 \pmod{8}$, $n \equiv 8 \pmod{27}$, $n \equiv 23 \pmod{125}$. Then $n, n + 1$, and $n + 2$ all have the property ($p$) because the factor 2, 3, and 5, respectively, has the exponent 2 in their prime factorizations.