Congruence condition fails for subgroups of order p^p and exponent p

Statement
Let $$p$$ be a prime number. Suppose $$P$$ is a group of prime power order for the prime $$p$$. Let $$p^k$$ be a prime power less than or equal to the order of $$P$$ and $$p^d$$ be a power of $$p$$ with $$d \le k$$. It is not necessary that the number of subgroups of order $$p^k$$ and exponent dividing $$p^d$$ is either zero or congruent to $$1$$ modulo $$p$$.

Opposite facts

 * Mann's replacement theorem for subgroups of prime exponent
 * Congruence condition on number of subgroups of given prime power order
 * Congruence condition on Sylow numbers
 * Congruence condition on number of subgroups of given prime power order and bounded exponent in abelian group

The case $$p = 2$$
Consider the dihedral group of order eight. The number of subgroups of order $$2^2 = 4$$ and exponent $$2^1 = 2$$ is equal to $$2$$, which is neither equal to zero or congruent to $$1$$ modulo $$2$$.

The case of odd $$p$$
Let $$p$$ be an odd prime, and let $$P$$ be the wreath product of groups of order p. Then, $$P$$ is a group of order $$p^{p+1}$$ and exponent $$p^2$$. Consider the subgroups of $$P$$ having order $$p^p$$ and exponent $$p$$. There are exactly two of these, both of them isomorph-free: the elementary abelian normal subgroup of order $$p^p$$, and the semidirect product of the commutator subgroup (which has order $$p^{p-1}$$) with the wreathing element of order $$p$$.