Congruence condition on number of subgroups of given prime power order and bounded exponent in abelian group

Statement
Let $$p$$ be a prime number and suppose $$P$$ is an abelian group of prime power order, i.e., a finite $$p$$-group. Suppose $$p^k$$ is a power of $$p$$ less than or equal to the order of $$P$$ and $$p^d$$ is a power of $$p$$ less than or equal to $$p^k$$.

Then, the number of subgroups of $$P$$ of order $$p^k$$ and exponent dividing $$p^d$$ is either equal to zero or congruent to $$1$$ modulo $$p$$.

Related facts

 * Congruence condition on number of subgroups of given prime power order
 * Jonah-Konvisser abelian-to-normal replacement theorem
 * Jonah-Konvisser elementary abelian-to-normal replacement theorem