There are counterexamples with $b=kp$ any multiple of $p$. Namely, for $a=1+k^pp^{p-1}$ we have
$$ a^p=1+\sum_{j=1}^p\binom{p}{j}(k^pp^{p-1})^j\equiv 1\pmod{k^pp^p}. $$
For example, $p=5$ and $k=3$ yields the congruence $151876^5\equiv 1\pmod{15^5}$.

Let us now assume that $b$ is not a multiple of $p$. Then $a^p\equiv 1\pmod{b^p}$ implies that either $a\equiv 1\pmod{b^p}$, or the multiplicative order of $a$ modulo $b^p$ equals $p$. In the latter case, we have $p\mid \varphi(b^p)=\varphi(b)b^{p-1}$, whence $p\mid\varphi(b)$, i.e. $b$ is a multiple of some prime $q\equiv 1\pmod{p}$. So in looking for a counterexample with $p\nmid b$, we may as well assume that $b$ is a prime congruent to $1$ mod $p$.

Here is a quick way to find a counterexample with $p\nmid b$ (for a full characterization see Vesselin Dimitrov's comment). Let $b$ be a prime congruent to $1$ mod $p$. Then $b^p$ is a prime power such that $\varphi(b^p)$ is divisible by $p$. Let $g$ be a primitive root mod $p$, and put $a:=g^{\varphi(b^p)/p}$. Then obviously $a^p\equiv 1\pmod{b^p}$, while $a\not\equiv 1\pmod{b^p}$.

**Added.** In retrospect the problem simply asks when does the group $(\mathbb{Z}/b^p\mathbb{Z})^\times$ have an element of order $p$. Formulated in this way, clearly a necessary and sufficient condition is $p\mid\varphi(b^p)$, which holds iff $b$ is divisible by $p$ or by a prime congruent to $1$ mod $p$.