It is $p-1$. I give an detailed explanation of why it is.     

##1. First, I give  an Corollary, it is used to prove the Elgamal Signature Verification.    
**Corollary** If $a$ and $p$ are relative prime integers, and $p$ is a prime integer,then $a^i \equiv a^j \pmod p$,where i and j are nonnegative integers, if and only if $i \equiv j \pmod {p-1}$.

This Corollary give the relationship of $p$ and $p-1$.     
  
One of the methods to prove it is based on  a  Theorem.        
The Theorem lies in 《elementary number thoery 6th》  **Theorem 9.2**. It can alse be used to prove the DSA.        

**Theorem 9.2** If $a$ and $n$ are relative prime integers with $n>0$,then $a^i \equiv a^j \pmod n$,where i and j are nonnegative integers, if and only if $i \equiv j \pmod {ord_n^a}$        
                   
Specially, when $p$ is a prime integer, ${ord_p^a}=\varphi(p)=p-1$ , so we prove the Corollary.             


hence:

$$i \equiv j \pmod {p-1} \Leftrightarrow a^i \equiv a^j \pmod p.$$

##2. Now, we review the Elgamal Signature, and prove it using the Corollary.        

###2.1 Key Generation for Elgamal Digital Signature    
>1. Choose a large prime $p$.
>2. Choose a primitive element $\alpha$ of $Z_p^*$.    
>3. Choose a random integer $d \in \{2,3, . . . , p-2\}$.    
>4. Compute$\beta = \alpha^d \pmod p$.     
>The public key is $k_{pub}=(p, \alpha, \beta)$, and private key $k_{pr}=d$


###2.2 Elgamal Signature Generation
The signing consists of two main steps: choosing a random value $k$, which forms an ephemeral private key, and computing the actual signature of $x$.        
>1. Choose a random ephemeral key $k \in \{0,1,2, . . . , p-2\}$ such that $gcd(k, p-1) = 1$.    
>2. Compute the signature parameters:
$$r \equiv \alpha^k mod p$$
$$s \equiv (x-d\cdot r)k^{-1} \pmod {p-1}$$

hence, the signature is: 
$$sig_{k_{pr}}(x,k)=(r,s)$$

###2.3 Elgamal Signature Verification

1. Compute the value
$$t \equiv \beta^r \cdot r^s \pmod p$$
2. The verification follows from:
$$
t\begin{cases}
\equiv \ \alpha^x \pmod p\ \ \ \  \Rightarrow  \ \ \ \ valid \ \ signature\\\\
\not \equiv \ \alpha^x \   \pmod p\ \ \ \  \Rightarrow  \ \ \ \  invalid \ \ signature
\end{cases}
$$


###2.4 Proof    

Prove that: 
$$ \alpha^x \equiv \beta^r \cdot r^s \pmod p$$
since:       
$$ \beta^r \cdot r^s \pmod p \equiv (\alpha^d)^r(\alpha^k)^s \pmod p \equiv \alpha^{d\cdot r+k\cdot s} \pmod p$$

So, we require that :     

$$\alpha^x \equiv \alpha^{d\cdot r+k\cdot s} \pmod p$$        

According the Corollary we give above, the relationship holds **if and only if**: 

$$ x \equiv d\cdot r+k\cdot s \pmod {p-1}$$   
hence, we get $s$:

$$s\equiv (x - d \cdot r)k^{-1} \pmod {p-1}$$

This is just the construction rule of the signature parameters s follows.               
The condition that $gcd(k, p-1) = 1$ is required since we have to invert the
ephemeral key modulo $p-1$ when computing s.    

##3. Conclusion    

From above discription, we know that why it is $p-1$, not $p$.        

## Reference     
1. 《elementary number theory 6th》
2. 《Understanding cryptography》