Relaxing the stack in a push down automata
I think the answer is no, even if we disallow unreachable states.

Consider the language $L$ given by the grammar $S ::= \varepsilon \mid abS \mid aSb$.  In this language, every word $w \in L$ has the same number of $a$'s and $b$'s, and every prefix of $w$ has at most as many $b$'s as $a$'s.  Also, if $w \in L$ is not empty, then $w$ starts with $a$ and has length $\ge 2$.

First PDA

Here's one PDA that can recognize that grammar.  It has two states $q_0,q_1$, and the following transitions:

$q_0 \stackrel{a,\epsilon\to \alpha}{\to} q_0$

$q_0 \stackrel{b,\alpha \to \epsilon}{\to} q_0$

$q_0 \stackrel{\epsilon, \epsilon \to \epsilon}{\to} q_1$.

The state $q_1$ is the accept state, and both states are reachable.

If you convert this to a NFA, then the NFA accepts the language $(a|b)^*$.

Second PDA

Here's another PDA that accepts the language $L$.  It has four states $q_0,q_1,q_2,q_3$, and transitions

$q_0 \stackrel{a,\epsilon \to \alpha}{\to} q_1$

$q_1 \stackrel{a,\epsilon\to \alpha}{\to} q_2$

$q_1 \stackrel{b,\alpha \to \epsilon}{\to} q_2$

$q_2 \stackrel{a,\epsilon\to \alpha}{\to} q_2$

$q_2 \stackrel{b,\alpha \to \epsilon}{\to} q_2$

$q_2 \stackrel{\epsilon, \epsilon \to \epsilon}{\to} q_3$

Here $q_0,q_3$ are the accepting states.  All states are reachable.

If you convert this to a NFA, it accepts the language $\varepsilon | a (a|b)^*$.