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Using the Mobius inversion formula, one may write

$$\Phi _n(X)= \prod _{d\mid n} (X^d-1)^{\mu (n/d)}.$$ We then get $$\frac{\Phi _n(p)}{p^{\phi (n)}}=\prod _{d \mid n} (1-\frac{1}{p^d})^{\mu (n/d)} . $$

[Edit] I did not think this would give a complete proof, but Fedja's comments below give the estimate in all cases. I give his proof: taking logs on both sides of the last equation we get $$\log \big (\frac{\Phi _n(p)}{p^{\phi (n)}}\big )= \sum _{d\mid n} \mu (\frac{n}{d})\log (1-\frac{1}{p^d})=- \sum _{d\mid n} \mu (\frac{n}{d}) \sum _{k\geq 1} \frac{1}{kp^{kd}}.$$ Now, for $0\leq x\leq 1/2$, it is proved in Fedja's comment below that the estimate $$0\leq \sum _{d\mid n}\mu (\frac{n}{d})x^d \leq x$$ holds. Taking $x=\frac{1}{p^k}$ and using this estimate in the above equality, we get the estimate
$$\mid \log \big( \frac{\Phi _n(p)}{p^{\phi (n)}}) \mid \leq \sum _{k\geq 1} \frac{1}{kp^k}=\log \big ( \frac{1}{1-\frac{1}{p} }\big ). $$ (We may assume that $\Phi _n(p)/p^{\phi (n)}\geq 1$; otherwise, the desired estimate is trivially true). Then the last inequality immediately implies $$\frac{\Phi _n(p)}{p^{\phi (n)}}\leq \frac{p}{p-1},$$ which is what was needed.

Using the Mobius inversion formula, one may write

$$\Phi _n(X)= \prod _{d\mid n} (X^d-1)^{\mu (n/d)}.$$ We then get $$\frac{\Phi _n(p)}{p^{\phi (n)}}=\prod _{d \mid n} (1-\frac{1}{p^d})^{\mu (n/d)} . $$

[Edit] I did not think this would give a complete proof, but Fedja's comments below give the estimate in all cases. I give his proof: taking logs on both sides of the last equation we get $$\log \big (\frac{\Phi _n(p)}{p^{\phi (n)}}\big )= \sum _{d\mid n} \mu (\frac{n}{d})\log (1-\frac{1}{p^d})=- \sum _{d\mid n} \mu (\frac{n}{d}) \sum _{k\geq 1} \frac{1}{kp^{kd}}.$$ Now, for $0\leq x\leq 1/2$, it is proved in Fedja's comment below that the estimate $$0\leq \mid \sum _{d\mid n}\mu (\frac{n}{d})x^d \mid \leq x$$ holds. Taking $x=\frac{1}{p^k}$ and using this estimate in the above equality, we get the estimate
$$\mid \log \big( \frac{\Phi _n(p)}{p^{\phi (n)}}) \mid \leq \sum _{k\geq 1} \frac{1}{kp^k}=\log \big ( \frac{1}{1-\frac{1}{p} }\big ). $$ (We may assume that $\Phi _n(p)/p^{\phi (n)}\geq 1$; otherwise, the desired estimate is trivially true). Then the last inequality immediately implies $$\frac{\Phi _n(p)}{p^{\phi (n)}}\leq \frac{p}{p-1},$$ which is what was needed.

Using the Mobius inversion formula, one may write

$$\Phi _n(X)= \prod _{d\mid n} (X^d-1)^{\mu (n/d)}.$$ We then get $$\frac{\Phi _n(p)}{p^{\phi (n)}}=\prod _{d \mid n} (1-\frac{1}{p^d})^{\mu (n/d)} . $$

[Edit] I did not think this would give a complete proof, but Fedja's comments below give the estimate in all cases. I give his proof: taking logs on both sides of the last equation we get $$\log \big (\frac{\Phi (p)}{p^{\phi (n)}}\big )= \sum _{d\mid n} \mu (\frac{n}{d})\log (1-\frac{1}{p^d})=- \sum _{d\mid n} \mu (\frac{n}{d}) \sum _{k\geq 1} \frac{1}{kp^{kd}}.$$ Now, for $0\leq x\leq 1/2$, it is proved in Fedja's comment below that the estimate $$0\leq \sum _{d\mid n}\mu (\frac{n}{d})x^d \leq x$$ holds. Taking $x=\frac{1}{p^k}$ and using this estimate in the above equality, we get the estimate
$$\mid \log \big( \frac{\Phi _n(p)}{p^{\phi (n)}}) \mid \leq \sum _{k\geq 1} \frac{1}{kp^k}=\log \big ( \frac{1}{1-\frac{1}{p} }\big ). $$ (We may assume that $\Phi _n(p)/p^{\phi (n)}\geq 1$; otherwise, the desired estimate is trivially true). Then the last inequality immediately implies $$\frac{\Phi _n(p)}{p^{\phi (n)}}\leq \frac{p}{p-1},$$ which is what was needed.

Using the Mobius inversion formula, one may write

$$\Phi _n(X)= \prod _{d\mid n} (X^d-1)^{\mu (n/d)}.$$ We then get $$\frac{\Phi _n(p)}{p^{\phi (n)}}=\prod _{d \mid n} (1-\frac{1}{p^d})^{\mu (n/d)} . $$

[Edit] I did not think this would give a complete proof, but Fedja's comments below give the estimate in all cases. I give his proof: taking logs on both sides of the last equation we get $$\log \big (\frac{\Phi _n(p)}{p^{\phi (n)}}\big )= \sum _{d\mid n} \mu (\frac{n}{d})\log (1-\frac{1}{p^d})=- \sum _{d\mid n} \mu (\frac{n}{d}) \sum _{k\geq 1} \frac{1}{kp^{kd}}.$$ Now, for $0\leq x\leq 1/2$, it is proved in Fedja's comment below that the estimate $$0\leq \sum _{d\mid n}\mu (\frac{n}{d})x^d \leq x$$ holds. Taking $x=\frac{1}{p^k}$ and using this estimate in the above equality, we get the estimate
$$\mid \log \big( \frac{\Phi _n(p)}{p^{\phi (n)}}) \mid \leq \sum _{k\geq 1} \frac{1}{kp^k}=\log \big ( \frac{1}{1-\frac{1}{p} }\big ). $$ (We may assume that $\Phi _n(p)/p^{\phi (n)}\geq 1$; otherwise, the desired estimate is trivially true). Then the last inequality immediately implies $$\frac{\Phi _n(p)}{p^{\phi (n)}}\leq \frac{p}{p-1},$$ which is what was needed.

Using the Mobius inversion formula, one may write

$$\Phi _n(X)= \prod _{d\mid n} (X^d-1)^{\mu (n/d)}.$$ We then get $$\frac{\Phi _n(p)}{p^{\phi (n)}}=\prod _{d \mid n} (1-\frac{1}{p^d})^{\mu (n/d)} . $$

[Edit] I did not think this would give a complete proof, but Fedja's comments below give the estimate in all cases. I give his proof: taking logs on both sides of the last equation we get $$\log \big (\frac{\Phi (p)}{p^{\phi (n)}}\big )= \sum _{d\mid n} \mu (\frac{n}{d})\log (1-\frac{1}{p^d})=- \sum _{d\mid n} \mu (\frac{n}{d}) \sum _{k\geq 1} \frac{1}{kp^{kd}}.$$ Now, for $0\leq x\leq 1/2$, it is proved in Fedja's comment below that the estimate $$0\leq \sum _{d\mid n}\mu (\frac{n}{d})x^d \leq x$$ holds. Taking $x=\frac{1}{p^k}$ and using this estimate in the above equality, we get the estimate
$$\mid \log \big( \frac{\Phi (p)}{p^{\phi (n)}}) \mid \leq \sum _{k\geq 1} \frac{1}{kp^k}=\log \big ( \frac{1}{1-\frac{1}{p} }\big ). $$ (We may assume that $\phi _n(p)/p^{\phi (n)}\geq 1$; otherwise, the desired estimate is trivially true). Then the last inequality immediately implies $$\frac{\Phi _n(p)}{p^{\phi (n)}}\leq \frac{p}{p-1},$$ which is what was needed.

Using the Mobius inversion formula, one may write

$$\Phi _n(X)= \prod _{d\mid n} (X^d-1)^{\mu (n/d)}.$$ We then get $$\frac{\Phi _n(p)}{p^{\phi (n)}}=\prod _{d \mid n} (1-\frac{1}{p^d})^{\mu (n/d)} . $$

[Edit] I did not think this would give a complete proof, but Fedja's comments below give the estimate in all cases. I give his proof: taking logs on both sides of the last equation we get $$\log \big (\frac{\Phi (p)}{p^{\phi (n)}}\big )= \sum _{d\mid n} \mu (\frac{n}{d})\log (1-\frac{1}{p^d})=- \sum _{d\mid n} \mu (\frac{n}{d}) \sum _{k\geq 1} \frac{1}{kp^{kd}}.$$ Now, for $0\leq x\leq 1/2$, it is proved in Fedja's comment below that the estimate $$0\leq \sum _{d\mid n}\mu (\frac{n}{d})x^d \leq x$$ holds. Taking $x=\frac{1}{p^k}$ and using this estimate in the above equality, we get the estimate
$$\mid \log \big( \frac{\Phi _n(p)}{p^{\phi (n)}}) \mid \leq \sum _{k\geq 1} \frac{1}{kp^k}=\log \big ( \frac{1}{1-\frac{1}{p} }\big ). $$ (We may assume that $\Phi _n(p)/p^{\phi (n)}\geq 1$; otherwise, the desired estimate is trivially true). Then the last inequality immediately implies $$\frac{\Phi _n(p)}{p^{\phi (n)}}\leq \frac{p}{p-1},$$ which is what was needed.