Using Branch if-else

Actually, you are using conditions so go ahead with branched if-else

const getDeck = () => {
      const suits = ["Spades", "Hearts", "Diamonds", "Clubs"];
      let cards = [];
    
      for (let s = 0; s < suits.length; s++) {
        for (let d = 0; d < 13; d++) {
          let displayVal;
          let val;
    
          if (d === 1) {
            displayVal = "Ace";
            val = 11;
          } else if (d >= 2 && d <= 10) {
            displayVal = d;
            val = d;
          } else if (d === 11) {
            displayVal = "Jack";
            val = 10;
          } else if (d === 12) {
            displayVal = "Queen";
            val = 10;
          } else if (d === 13) {
            displayVal = "King";
            val = 10;
          } else {
            displayVal = "";
            val = 0;
          }
          cards.push({
            val,
            displayVal,
            suit: suits[s],
          });
        }
      }
      return cards;
    };
    
    const deck = getDeck();
    console.log(deck);
    console.log(`Deck length equals 52? ${deck.length === 52}`);

Using Switch case

const getDeck = () => {
    const suits = ['Spades','Hearts','Diamonds','Clubs'];
    let cards = [];
  
    for(let s=0; s < suits.length; s++){
  
     for(let d = 0; d < 13; d++){
      let displayVal;
      let val;
  
       switch (true) {
        case (d === 1):
          displayVal = "Ace";
          val = 11;
          break;
        case (d >=2 && d <= 10):
           displayVal = d;
           val = d;
           break;
          case (d === 11):
            displayVal = "Jack"
            val = 10;
            break;
          case (d === 12):
            displayVal = "Queen"
            val = 10;
            break;
          case (d === 13):
            displayVal = "King"
            val = 10;
          default:
            displayVal =""
            val =0
       }
      cards.push({
      val,
      displayVal,
      suit: suits[s]
    } ); 
  
     }
  
   }
   return cards;
  }
  
  const deck = getDeck();
  console.log(deck);
  console.log(`Deck length equals 52? ${deck.length === 52}`);

There is an alternative way / approach that may be used here.

Code Snippet

const getDeck = () => {
  const suits = ['Spades','Hearts','Diamonds','Clubs'];
  const getValueOf = x => x === 1 ? 11 : x > 10 ? 10 : x;
  const displayValMap = {1: 'Ace', 11: 'Jack', 12: 'Queen', 13: 'King'};
  return suits.map(
    suit => (
      [
        ...Array(13)
        .keys()
      ]
      .map(x => x + 1)
      .map(d => ({
        val: getValueOf(d),
        displayVal: d in displayValMap
          ? displayValMap[d]
          : d,
        suit
      }))
    )
  ).flat();
}

const deck = getDeck();
console.log(`Deck length equals 52? ${deck.length === 52}`);
console.log(deck);

Explanation

• Set-up ways to get val and displayVal (in above snippet, getValueOf - a one-line method; and displayValMap an object, is used)
• Iterate through the suits
• Use ... spread-operator to generate an array of length 13, then get the keys of the array.
• Use .map to make the generated array start from 1, instead of 0
• For each element in the array, create an object with val, displayVal and suit props
• And, finally, use .flat() to return all objects into a single array instead of nested-arrays.

Another way may be:

const getDeck = () => ['Spades','Hearts','Diamonds','Clubs'].map(
    suit => (
      [
        ...Array(13)
        .keys()
      ]
      .map(d => ({
        val: (d === 0 ? 10 : d > 9 ? 9 : d) + 1,
        displayVal: {0: 'Ace', 10: 'Jack', 11: 'Queen', 12: 'King'}[d] || d + 1,
        suit
      }))
    )
).flat();

console.log(getDeck().length);
console.log(getDeck());