To reiterate Stanley Yao Xiao's comment, one should not expect that $p-1$ has any more or less prime factors than a typical integer of size $p$. For an arbitrary integer $n$, the bound $$ \omega(n) \ll \frac{\log n}{\log\log n} $$ is standard (see Hardy and Wright's "An Introduction to the Theory of Numbers" for a proof of this). Here $\omega(n)$ denotes the number of distinct prime factors of $n$.

It is known that integers of the form $p-1$ have about the same number of divisors as typical integers $n$ on average. The Titchmarsh divisor problem states that $$ \sum_{p\leq x} \tau(p-1) \sim C x $$ for some explicit constant $C$, where $\tau(n)$ denotes the number of positive divisors of $n$. Stated differently, we may write $$ \frac{1}{\pi(x)} \sum_{p\leq x} \tau(p-1) \sim C \log x $$ by the Prime Number Theorem, and so $p-1$ has $\asymp \log x$ prime factors on average for $p\leq x$, the same as for all integers up to $x$ (up to a constant factor).

To reiterate Stanley Yao Xiao's comment, one should not expect that $p-1$ has any more or less prime factors than a typical integer of size $p$. For an arbitrary integer $n$, the bound $$ \omega(n) \ll \frac{\log n}{\log\log n} $$ is standard (see Hardy and Wright's "An Introduction to the Theory of Numbers" for a proof of this). Here $\omega(n)$ denotes the number of distinct prime factors of $n$.

It is known that integers of the form $p-1$ have about the same number of divisors as typical integers $n$ on average. The Titchmarsh divisor problem states that $$ \sum_{p\leq x} \tau(p-1) \sim C x $$ for some explicit constant $C$, where $\tau(n)$ denotes the number of positive divisors of $n$. Stated differently, we may write $$ \frac{1}{\pi(x)} \sum_{p\leq x} \tau(p-1) \sim C \log x $$ by the Prime Number Theorem, and so $p-1$ has $\asymp \log x$ divisors on average for $p\leq x$, the same as for all integers up to $x$ (up to a constant factor).